GIFT   OF 
MICHAEL  REESE 


SLIDE-VALVES. 

A  BOOK  FOR  PRACTICAL  MEN 

+ 

ON   THE 

PRINCIPLES  AND   METHODS   OF  DESIGN; 

WITH 

AN  EXPLANATION  OF  THE  PRINCIPLES 
OF  SHAFT-GOVERNORS. 


BY 

C.   W.   MAcCORD,  JR.,   M.E. 


FIRST   EDITION. 
FIRST    THOUSAND. 


NEW   YORK: 

JOHN   WILEY   &   SONS. 

LONDON:   CHAPMAN  &  HALL,  LIMITED. 

1897. 


Copyright,  1897, 

BY 

C.  W.  MAcCORD,  JR. 


ROBERT  DRUMMOND.   KLECTROTYPER   AND   PRINTER,   NEW  YORK. 


PREFACE. 


A  SERIES  of  articles  entitled  "Valve-Gears"  was  pub- 
lished by  the  author  in  Power  during  1895  and  1896.  The 
object  of  these  papers  was  to  put  the  principles  of  design 
of  slide-valves  in  a  practical  form  for  practical  men.  This 
book  contains  the  subject-matter  of  those  papers  entirely 
revised  and  rearranged,  with  a  complete  new  set  of  cuts. 

C.  W.  MACCORD  JR. 

AUBURN,  N.  Y., 

July  4,  1897. 

JT*Se^    ~     "-'"^r^A^ 

iii 


CONTENTS. 


CHAPTER   I. 

PAGE 

General  Principles I 

CHAPTER   II. 
Crank,  Connecting-Rod,  and  Eccentric-Rod 17 

CHAPTER   III. 
Valve  Diagrams 27 

CHAPTER   IV. 
Steam  and  Exhaust  Ports — Bridges— Steam-Pipes 44 

CHAPTER   V. 
General  Problems  in  Valve  Design 54 

CHAPTER   VI. 
Rockers  and  Bell-Cranks 62 

CHAPTER   VII. 
Design  of  a  Plain  D-Valve 68 

CHAPTER  VIII. 

Valve  and  Eccentric  Rods,  Steam-Chest , 78 

v 


vi  CONTENTS. 

CHAPTER   IX. 

PAGE 

Design  of  an  Allen  or  " Trick"  Valve 82 

CHAPTER  X. 
Design  of  a  Double-Ported  Valve • 9° 

CHAPTER  XI. 
Valve-Setting 99 

CHAPTER  XII. 
Shaft-Governors—General  Principles  and  Types 108 

CHAPTER  XIII. 
Shaft-Governors—Analysis  of  Principles • 124 


SLIDE-VALVES. 


CHAPTER    I. 
GENERAL   PRINCIPLES. 

FlG.  I  shows  a  cylinder  which  is  entirely  empty,  except 
for  the  piston  shown;  the  latter  being  free  to  slide  in  the 
cylinder  under  the  influence  of  any  force.  Suppose,  for 


PlG.l 


example,  that  steam  were  admitted  to  the  space  A  at  the 
left.  The  natural  result  would  be  that  the  piston  would 
move  over  to  the  right-hand  end,  as  shown  in  Fig.  2.  Now, 


FIG.  2 


in   order  to   bring  the  piston  back  to   the  position  shown  in 
Fig.  I,  a  force  must  be  applied  to  the  right  of  the  piston. 


2  SLIDE-VALVES. 

Suppose  steam  were  admitted  there.  This  would  not  suffice 
to  move  the  piston,  because  the  space  A  is  filled  with  steam 
and  the  pressures  on  the  two  sides  of  the  piston  would 
balance  each  other.  It  therefore  follows  that  the  mere 
admission  of  steam  to  the  ends  of  the  cylinder  in  alternation 
is  not  enough  to  cause  the  piston  to  reciprocate,  or  move 
backward  and  forward ;  but,  in  addition,  some  means  must  be 
provided  for  emptying  or  exhausting  the  steam  from  the 
cylinder  as  soon  as  it  has  done  its  work.  If  this  exhausting 
is  accomplished,  the  steam-pressure  will  then  move  the  piston 
in  the  desired  way. 

In  Fig.  3  is  represented  the  simplest  contrivance  designed 
to   regulate   the   admission   and   exhaust  of  steam  from   the 


FIG.  3 


cylinder.  The  space  S  is  called  the  " steam-chest,"  and  is 
in  direct  communication  with  the  boiler,  so  that  it  is  at  all 
times  full  of  steam  at  boiler-pressure.  The  passages  P  and 
P'  lead  to  the  right-  and  left-hand  ends,  respectively,  of  the 
cylinder,  and  are  called  " steam-ports  "  or,  more  often, 
"porf-s."  The  piece  Fis  the' 'valve,"  and  it  is  free  to  slide 
in  a  horizontal  direction  only,  or  parallel  with  the  "  line  of 


GENERAL   PRINCIPLES.  5 

stroke"  of  the  piston.  The  surface  of  the  main  casting  in 
which  the  ports  terminate  is  called  the  " valve-seat,"  and 
the  lower  surface  of  the  valve,  which  is  in  contact  with  the 
seat,  is  called  the  " valve-face."  The  valve  is  moved  by 
means  of  the  "va'lve-rod,"  VR,  which  passes  through  a 
" stuffing-box  "  at  the  left  of  the  steam-chest;  the  object  of 
the  stuffing-box  being  to  prevent  the  leakage  of  steam.  The 
piston,  A,  with  its  "piston-rod,"  does  not  require  any  partic- 
ular description.  The  passage  E  in  the  valve-seat  is  called 
the  "exhaust-port,"  and  it  is  in  direct  communication  with 
the  atmosphere  at  all  times. 

The  valve  as  shown  is  separated  entirely  frcm  the  "valve- 
gear"  which  operates  it.  This  is  done  because  a  clearer 
understanding  of  the  subject  can  be  had  by  first  studying 
thoroughly  the  movements  of  the  valve  and  then  considering 
the  gear  required  to  produce  these  movements. 

The  operation  of  this  valve  is  as  follows: 

In  Fig.  3  the  valve  just  covers  both  ports,  and  the  steam 
cannot  reach  either  end  of  the  cylinder.  Then  the  valve 


FIG.  4 


moves  slightly  to  the  left,   assuming  the  position  shown  in 
Fig.  4,  when  the  right-hand  port  is  uncovered,  and  the  steam 


4  SLIDE-VALVES. 

naturally  follows  the  course  indicated  by  the  curved  arrows, 
thus  getting  on  the  right  of  the  piston  and  forcing  it  over  to 
the  left,  as  shown  by  the  straight  arrow.  When  the  piston 
reaches  the  middle  of  its  stroke  the  valve  is  at  the  extreme 
left  of  its  stroke,  or  travel,  and  begins  to  move  in  the  oppo- 
sition direction;  this  being  shown  in  Fig.  5.  When  the 
piston  reaches  the  extreme  left-hand  position  the  valve  is 


FIG.  5 

again  "  central  or  in  the  middle  position,  and  just  covers 
both  ports,  as  shown  in  Fig.  6.  The  valve  continues  to 
move  to  the  right,  thus  admitting  steam  to  the  left-hand  end 
of  the  cylinder,  as  shown  in  Fig.  7.  At  the  same  time,  it 
will  be  noticed,  the  cavity  E'  in  the  valve  is  over  the  right- 
hand  steam-port,  thus  permitting  steam  to  escape  from  the 
right-hand  side  of  the  piston  through  the  exhaust-port  E,  as 
shown  by  the  arrows.  The  exhaust-port  leads  to  the  atmos- 
phere. The  piston  now  reverses  its  stroke,  and  when  in  the 
middle  of  its  stroke  the  valve  is  over  at  the  extreme  right- 
hand  end  of  its  travel,  as  shown  in  Fig.  8.  The  piston  con- 
tinues to  move  to  the  right,  while  the  valve  reverses  and 
moves  to  the  left,  and  the  original  condition,  shown  in 


GENERAL  PRINCIPLES. 


5 


Fig.  3,  is  once  more  attained.     The  relative  movements  of 
the  piston  and  valve  are  then,  repeated.     When  the  valve  is 


FIG.  6 


again  over  at  the  left,  as  shown  in  Figs.  4  and  5,  the  steam 
at  the  left  of  the  piston  is  exhausted  through  P'  and  £. 


v  it 


FIG.  7 

It  should  be  at  once  apparent  that  the  motion  of  the  valve 
bears  a  fixed  relation  to  the  motion  of  the  piston,  and  should 


SLIDE-VALVES. 


be  derived  from  it  in  some  way.     When  the  valve  is  in  its 
central  position  the  piston  is  at  either  one  end  or  the  other  of 


FIG.  8 


its  stroke.     That  is,  the  valve  is  either  a  half-stroke  ahead  of 
or  behind  the  piston. 

In  every  engine  the  reciprocating  or  "to-and-fro  "  motion 
of  the  piston  is  converted  into  a  rotary  motion  of  the  fly- 
wheel. Fig.  9  shows  the  simplest  possible  contrivance  for 
effecting  this  conversion;  and  at  the  same  time  it  shows  the 
simplest  form  of  apparatus  for  deriving  the  motion  of  the 
valve  from  the  stroke  of  the  piston.  The  center  of  the 
crank-shaft,  on  which  the  fly-wheel  is  keyed,  is  at  <9,  and  the 
crank-pin  is  at  C.  The  crank  itself  is  omitted  for  the  sake  of 
clearness.  The  piston-rod,  R,  terminates  in  a  slotted  cross- 
head,  in  which  the  crank-pin  fits  and  is  free  to  slide.  With 
this  gear,  as  the  piston  moves  to  and  fro,  the  crank  rotates, 
and  the  crank-pin  slides  up  and  down  in  the  slotted  cross-head. 
It  will  be  noticed  that  when  the  positions  of  the  piston-rod, 
crank-pin,  and  shaft  are  as  shown  in  the  figure — that  is,  when 
their  centers  are  all  in  the  same  straight  line, — there  is  no 
rotative  effort  on  the  crank-pin,  the  entire  pressure  on  the 


GENERAL   PRINCIPLES.  7 

piston  being  expended  in  an  attempt  to  crush  the  crank-pin 
or  shaft.  At  the  other  end  of  the  stroke  a  similar  state  of 
affairs  occurs.  These  two  points  are  called  the  "dead- 
points"  or  " dead-centers."  If  an  engine,  when  stopped, 
gets  on  either  dead-center,  it  is  necessary  to  pry  it  off  before 


FIG.  9 


it  can  be  started.     When  the  engine  is   on  the  center  the 
piston  is  at  the  end  of  its  stroke. 

Another  pin  is  secured  to  the  shaft  just  90°,  or  a  quarter 
circumference,  away  from  C,  as  shown  at  E.  If,  therefore, 
the  valve  be  driven  by  E,  it  will  always  be  a  half-stroke  away 
from  the  piston.  The  pin  E  is  called  the  "eccentric-pin," 
or  the  "  eccentric,"  and  another  slotted  cross-head  is  em- 
ployed to  communicate  the  motion  of  E  to  the  valve-rod, 
VR,  through  the  "rocker,"  or  "  rock-shaft,"  L.  This 
rocker  is  pivoted  at  £>,  which  is  some  point  on  the  frame  of 
the  engine;  and,  in  the  case  shown,  the  arms  are  equal,  so 
that  the  motion  of  the  valve  is  the  same  in  amount  as  if  the 
valve-rod  were  fastened  to  the  end  of  the  eccentric-rod ;  but 
it  is  opposite  in  direction,  or  the  rocker  reverses  the  move- 


8  SLIDE-VALVES. 

ment  of  the  valve.  A  rocker  is  not  an  indispensable  adjunct, 
but  it  is  used  when  the  line  of  travel  of  the  valve  is  above  or 
below  the  line  of  stroke  of  the  engine.  The  type  shown  in 
the  cut  is  only  one  of  many,  some  of  which  will  be  described 
later,  and  is  taken  on  account  of  its  simplicity. 

This  device  answers  the  two  requirements  mentioned; 
for  the  motion  of  the  valve  is  obviously  derived  from  the 
stroke  of  the  piston,  and,  as  the  angle  between  the  crank  and 
the  eccentric  is  90°,  it  is  evident  that  when  the  piston  is  at 
one  end  of  its  stroke,  as  it  must  be  in  Fig.  9,  the  rocker  L  is 
vertical  and  the  valve  is  consequently  in  the  middle  position. 

The  distance  OE  from  the  center  of  the  crank-shaft  to  the 
center  of  the  eccentric-pin  is  called  the  "throw"  of  the 
eccentric,  or  simply  the  "  eccentricity,"  and  a  momentary 
inspection  of  the  figure  will  show  that  the  travel  of  the  valve 
is  equal  to  twice  the  eccentricity,  or  to  the  diameter  of  the 
circle  in  which  the  eccentric-pin  travels.  Similarly,  the 
length  of  the  crank,  OC,  is  often  called  the  "throw  "  of  the 
crank,  and  it  is  equal  to  one-half  the  stroke  of  the  engine. 
It  will  be  noticed  that  the  travel  of  the  valve  is  much  less 
than  that  of  the  piston.  It  is  made  so  in  order  to  decrease 
the  power  required  to  move  the  valve;  for  the  friction 
between  the  valve  and  the  valve-seat  varies  directly  with  the 
distance  through  which  the  valve  has  to  be  moved. 

This  type  of  valve-gear  has  two  serious  drawbacks:  The 
steam  is  not  used  expansively,  and  there  is  no  "compres- 
sion "  or  cushioning  of  the  steam  at  the  end  of  the  stroke  to 
take  up  the  momentum  of  the  moving  parts,  to  avoid 
"  pounding."  Both  of  these  faults  are  due  to  the  fact  that 
steam  is  admitted  to  the  cylinder  from  the  moment  that  the 
piston  leaves  one  end  of  the  cylinder  until  it  reaches  the  other 
end,  or,  in  other  words,  "steam  follows  full  stroke,"  a  fact 
which  should  be  plain  from  the  foregoing  text.  The  exhaust 
is  also  open  during  the  entire  stroke. 

If  steam  were  "cut  off"  from   the  cylinder  some  time 


GENERAL   PRINCIPLES.  9 

before  the  piston  reached  the  end  of  its  stroke,  the  expansive 
force  of  the  steam  would,  in  combination  with  the  reduction 
of  pressure  in  front  of  the  piston  due  to  the  exhaust,  carry 
the  piston  along  to  the  end  of  the  stroke.  There  is  great 
gain  in  economy  in  so  using  steam  expansively,  but  it  lies 
without  the  province  of  this  work  to  enter  upon  its  proof. 
Similarly,  if  the  port  were  closed  to  exhaust  a  short  time 
before  the  piston  reached  the  end  of  its  stroke,  a  certain  vol- 
ume of  steam  would  be  imprisoned  and  compressed  by  the 
advancing  piston,  thus  acting  as  a  cushion  to  overcome  the 
momentum  of  the  moving  parts. 

Steam  may  be  cut  off  from  the  cylinder  at  any  desired 
point  in  the  stroke  of  the  piston  by  adding  to  the  outside  of 
the  valve  a  piece  equal  to  the  distance  that  the  valve  has 
moved  from  its  central  position  when  the  piston  is  at  that 
point  of  the  stroke.  This  is  shown  in  Fig.  10,  which  is  a 


center-line  sketch  of  the  valve-gear  shown  in  Fig.  9,  with  the 
addition  of  the  valve  and  ports  at  the  end  of  the  valve-rod. 
Suppose,  for  example,  that  it  is  desired  to  cut  off  at  three- 
fourths  stroke — that  is,  when  the  piston  has  completed  three 


10  SLID  E-  VA  L  VES. 

fourths  of  its  travel  toward  one  end  of  the  cylinder  the  port 
shall  be  closed  to  steam. 

FIRST.  Find  the  position  of  the  crank-pin  when  the  piston 
is  at  the  point  where  cut-off  is  required.  This  is  done  by 
describing,  about  O  as  a  center,  a  circle  with  a  radius  equal 
to  the  length  of  the  crank ;  then  take  TF  equal  to  |  of  A  T, 
and  draw  a  perpendicular  to  A  T  from  F  until  it  cuts  the 
crank-circle  at  C,  the  required  crank-pin  position ;  and  OC  is 
the  "  crank-line,"  or  line  representing  the  position  of  the 
center-line  of  the  crank.  This  is  true  because,  with  the 
slotted  connection,  the  horizontal  movement  of  the  crank-pin 
is  the  same  as  that  of  the  piston. 

SECOND.  Locate  the  eccentric-pin  position  corresponding  to 
the  determined  crank-pin  position.  This  is  done  by  drawing 
OE,  the  eccentric  line,  perpendicular  to  OC,  because  the 
angle  between  the  crank  and  eccentric  is  90°,  and  describing 
a  circle  about  O  as  a  center,  and  with  a  radius  equal  to  the 
eccentricity.  This  circle  cuts  OE  at  E,  which  is  the  required 
position  of  the  eccentric-pin. 

THIRD.  Locate  the  position  of  the  end  of  the  eccentric-rod. 
To  do  this,  it  is  only  necessary  to  drop  a  perpendicular  ED 
from  E  to  the  line  of  stroke. 

FOURTH.  Draw  in  the  rest  of  the  figure.  This  is  simply 
done  by  taking  the  dimensions  of  the  valve-rod,  eccentric-rod, 
rocker,  etc.,  from  Fig.  9. 

Now,  the  "displacement  of  the  valve,"  or  the  distance  it 
has  moved  from  its  central  position,  is  OR  or  /,  and  it  is  evi- 
dent that  the  valve  must  move  that  distance  to  the  right 
before  the  right-hand  port  is  closed.  If  a  piece  of  the  length 
/  is  added  to  the  outside  of  the  valve,  as  shown  by  the  dotted 
lines,  the  port  will  be  closed  with  the  gear  in  the  position 
indicated,  and  the  desired  result  will  be  obtained  for  the 
right-hand  end  of  the  stroke. 

The  angle  which  the  crank-line  at  any  time  makes  with 
the  line  of  stroke  is  called  the  ''crank-angle  ";  and  with  the 


GENERAL   PRINCIPLES.  II 

slotted  connection  it  is  evident  that  for  the  same  piston 
position  on  the  forward  and  return  strokes  the  crank-angles 
are  equal.  Consequently  the  valve  displacements  are  equal, 
and,  taking  the  case  just  solved,  it  is  only  necessary  to 
lengthen  the  valve  on  the  other  end  by  the  amount  /  in  order 
to  secure  cut-off  at  the  same  point  on  the  return  stroke. 

The  end  of  the 'cylinder  nearest  the  crank  is  called  the 
"  crank  end,"  the  other  being  called  the  "  head  end."  The 
stroke  in  which  the  piston  approaches  the  crank-shaft  is  called 
the  " forward  stroke";  and  the  stroke  in  which  it  recedes 
from  the  crank-shaft  is  called  the  "return  stroke." 

The  length  /which  is  added  is  called  the  "outside  lap  " 
or  "  steam-lap,"  because  it  is  on  the  outside  of  the  valve  and 
controls  the  admission  of  live  steam.  It  is  defined  as  the 
amount  which  the  valve  projects  over  the  outside  edge  of  the 
steam-port  when  the  valve  is  in  its  central  position,  as  shown 
in  Fig.  ii.  It  is  very  evident  that  if  the  valve  originally  just 


I,  I,  =  outside  laps. 

i,  t,  =  inside  laps. 

6,  b,=  bridges.  PIG.  11 

e,     =  exhaust  port. 

p,  p,=steam  ports. 

covered  the  port  when  in  its  central  position,  a  piece  of  length 
/  would  project  just  that  amount  over  the  edge  of  the  port. 

The  compression  of  steam  may  be  secured  by  adding  a 
length  /'  to  the  inside  of  the  valve,  the  value  of  i  being 
determined  in  the  same  manner  that  /was  found;  that  is.  by 
making  it  equal  to  the  distance  which  the  valve  would  have 
to  travel  in  order  to  close  the  port,  with  the  piston  in  the 


1 2  SLIDE-  VA  L  VES. 

required  position.  This  addition  is  called  the  "inside  lap'* 
or  "exhaust-lap,"  for  obvious  reasons.  It  maybe  defined 
as  the  amount  which  the  valve  projects  over  the  inside  edge 
of  the  port,  with  the  valve  central,  as  in  Fig.  n. 

The  partition  which  divides  the  exhaust  from  the  steam- 
port  is  called  the  bridge.  The  inside  lap  therefore  rests  on 
the  bridge. 

A  difficulty  due  to  the  introduction  of  outside  lap  is  at 
once  apparent.  The  admission  of  steam  is  made  late  by  just 
the  amount  /.  That  is,  the  valve  must  move  a  distance  / 
from  its  middle  position  before  the  port  is  opened  to  steam, 
or  the  eccentric  must  turn  through  the  corresponding  angle. 
This  may  be  remedied  by  changing  the  angle  between  the 
crank  and  eccentric,  moving  the  eccentric  ahead — that  is,  in 
the  direction  in  which  the  engine  is  to  run — an  angle  corre- 
sponding to  Z\  That  is,  in  Fig.  10,  through  an  angle  EOP 
making  the  eccentric  position  OE*.  This  will  make  the 
admission  occur  sooner,  but  it  will  also  make  the  cut-off 
occur  earlier,  so  that  the  result  will  be  different  from  the 
three-quarter  cut-off  desired,  as  shown  in  Fig.  12,  where  E* 
is  the  eccentric  position  with  the  port  closed,  and  OC  the 
corresponding  crank-position.  It  is  not  necessary  to  show 
how  this  three-quarter  cut-off  could  be  obtained,  as  the 
general  method  has  been  indicated,  and  this  type  of  gear  is 
only  useful  as  a  mode  of  explanation.  By  increasing  the  angle 
between  the  crank  and  eccentric,  the  opening  of  the  port  to 
exhaust  will  be  made  earlier — that  is,  it  will  occur  before  the 
end  of  the  stroke,  in  the  simple  valve  with  outside  lap  but  no 
inside  lap;  and  the  compression  will  occur  before  the  end  of 
the  return  stroke.  The  inside  lap,  if  added,  need  be  much 
smaller  than  the  outside  lap.  Early  exhaust  opening,  pro- 
vided it  does  not  occur  earlier  than  seven-eighths  stroke,  does 
no  harm,  as  it  simply  serves  to  reduce  the  pressure  against 
which  the  piston  must  work  on  its  return  stroke. 

The  angle  between  the  crank  and  the  eccentric  has  now 


GENERAL   PRINCIPLES.  13 

been  changed  from  90°  to  COE*  by  the  advance  of  the  eccen- 
tric, and  the  amount  of  the  change  is  called  the  "angular 
advance";  and  angular  advance  is  therefore  defined  as  the 


FIG. 12 

difference  between  90°  and  the  angle  between  the  crank  and 
eccentric. 

The  rocker  affects  very  materially  the  angle  between  the 
crank  and  eccentric.  With  the  device  shown,  the  horizontal 
motion  of  the  upper  arm  is  directly  opposite  to  that  of  the 
lower  arm.  If,  in  the  simple  device  shown  in  Fig.  9,  the 
eccentric-rod  were  fastened  direct  to  the  valve-rod,  and  the 
rocker  in  consequence  eliminated,  it  would  be  necessary  to 
move  the  eccentric-pin  around  to  E' ',  which  is  180°  away  from 
E,  in  order  to  secure  the  same  movement  of  the  valve.  If, 
with  the  eccentric  in  this  new  position,  outside  lap  were 
added  to  the  valve,  it  would  be  necessary  to  move  the  eccen- 
tric ahead — in  the  direction  in  which  the  engine  is  to  run — to 
secure  the  admission  of  steam  at  the  beginning  of  the  stroke, 
just  as  before.  But  in  this  case  it  would  tend  to  increase 
the  angle  between  the  crank  and  eccentric,  while  with  the 
rocker  it  resulted  in  a  decrease  of  the  angle.  In  the  case  of 


14  SLIDE-  VA  L  VES. 

direct  connection  the  eccentric  precedes  the  crank,  while  with 
the  reversing  rocker  the  eccentric  follows.  The  definition  of 
angular  advance  may  therefore  be  stated  as  follows: 

li Angular  advance"  is  the  difference  between  the  angle 
between  the  crank  and  eccentric  arid  90°. 

When  the  right  angle  is  the  greater  the  angle  becomes 
the  "angle  of  follow,"  or  the  eccentric  follows  the  crank. 

When  the  right  angle  is  the  less,  the  angle  becomes  the 
"angle  of  advance,"  or  the  eccentric  precedes  the  crank. 

This  affords  a  ready  means  of  determining  whether  a 
rocker  is  used. 

With  this  type  of  valve-gear  the  valve  displacement  for  a 
given  piston  position  is  found  by  drawing  the  perpendicular 
from  the  eccentric-pin  to  the  line  of  stroke,  as  ER,  Fig.  10. 
The  displacement  from  the  end  of  its  travel  is  RS,  and  from 
the  central  position  is  OR.  The  latter  is  the  one  usually 
employed.  It  must  be  remembered  that,  on  the  contrary, 
"piston  displacement  "  is  the  amount  the  piston  has  moved 
from  the  beginning  of  its  stroke,  and  is  usually  expressed  by 
the  fraction  of  the  stroke  completed. 

It  has  been  pointed  out  that  setting  the  eccentric  ahead 
makes  all  the  events  of  the  stroke  earlier.  This  may  be  car- 
ried to  such  an  extreme  as  to  reverse  the  direction  of  rotation 
of  the  crank.  This  may  be  explained  by  the  use  of  Fig.  13. 

Let  CO  be  the  crank  and  OE  the  eccentric,  with  the 
engine  on  the  head-end  dead-center,  running  in  the  direction 
shown  by  the  arrow.  The  relative  positions  of  crank  and 
eccentric  are  taken  from  Fig.  12.  At  this  stage  the  valve 
would  commence  to  move  to  the  left,  opening  the  head-end 
port.  But  suppose  that  the  eccentric-pin  were  suddenly 
slipped  around  to  E1,  the  angle  COE1  being  equal  to  COE. 
The  valve  displacement  would  remain  the  same,  so  that  the 
right-hand  port  would  be  just  covered.  But  if  the  engine 
continues  to  move  in  the  direction  of  the  arrow,  the  valve 
will  move  to  the  right,  with  the  eccentric  at  E\  thus  opening" 


GENERAL  PRINCIP 


the  right-hand  or  head-end  port  to  exhaust  and  the  crank-end 
port  to  steam.  The  difference  between  this  motion  of  the 
valve  and  that  obtained  with  the  eccentric  at  E  is  shown  in 
the  figure.  The  piston  is  moving  toward  the  left  in  both 
cases;  but  with  the  head-end  port  exhausting  and  the  crank- 


FIG.  13 


ECCENTRIC  AT  E. 


end  port  admitting  steam,  as  shown  by  the  upper  valve,  there 
soon  comes  a  time. when  the  pressure  at  the  left  of  the  piston 
overcomes  the  momentum  of  the  moving  parts  and  forces 
the  piston  to  reverse  its  movement.  This  carries  the  crank  in 
the  opposite  direction.  The  engine  will  continue  to  run  in 
this  reversed  direction,  as  E1  is  as  much  behind  the  crank  in 
the  new  motion  as  E  was  behind  the  old. 

It  will  be  noticed  that  in  the  first  case,  with  the  eccentric 
at  Ey  the  crank  goes  above  the  line  of  stroke  on  the  forward 
stroke.  The  engine  is  said  to  ''run  over."  When  reversed, 
the  crank  goes  below  the  line  of  stroke  on  the  forward  stroke, 
and  the  engine  "runs  under." 

The  results  obtained  from  Fig.  13  may  be  stated  as 
follows : 


1 6  SLIDE-  VA  L  VES. 

To  reverse  an  engine,  it  is  only  necessary  to  slip  the  eccen- 
tric-pin around  on  the  shaft  until  it  is  on  the  other  side  of  the 
crank-pin,  making  the  angular  distance  between  the  crank 
and  eccentric  the  same  that  it  was  at  first. 

In  this  connection  it  is  to  be  noted  that  it  is  not  correct 
to  slip  the  eccentric  around  1 80°  on  the  shaft,  until  it  is 
directly  opposite  its  original  position.  This  condition  is 
shown  by  OE\  and  it  is  at  once  apparent  that  that  will  make 
all  the  events  of  the  stroke  late  by  the  angle  E'OE\  Of 
course,  if  the  angle  between  the  crank  and  eccentric  were 
90°,  it  would  be  correct  to  slip  it  around  180°,  and  that  is 
the  only  case  in  which  it  would  be  right. 


CHAPTER    II. 
CRANKS,   CONNECTING-RODS,  AND   ECCENTRICS. 

THE  slotted  cross-head  form  of  connection  is  but  seldom 
used,  on  account  of  certain  mechanical  difficulties  which  it  is 
unnecessary  to  describe  here.  The  commonest  means  of 
converting  the  reciprocating  motion  of  the  piston  into  the 
rotary  motion  of  the  crank  is  shown  in  Fig.  14.  The  piston- 


FIG.  14 


rod,  B,  terminates  in  the  cross  head,  H,  to  which  it  is  rigidly 
secured,  the  cross-head  sliding  between  the  guides  G  G.  The 
crank-shaft,  S,  carries  the  crank  or  pitman,  M,  on  which  is 
the  crank-pin,  C.  The  connecting-rod,  R,  joins  the  crank- 
pin  with  the  wrist-pin,  W,  of  the  cross-head,  the  connecting- 
rod  being  provided  with  bearings  in  which  these  pins  fit,  so 
that  the  rod  may  alter  its  position  and  still  keep  the  distance 
between  the  pins  sensibly  constant,  provisions  being  made  in 
various  ways  for  taking  up  the  wear. 

Very  frequently — in  fact  always — the  eccentricity  is  less 
than  the  radius  of  the  crank-shaft.      If,  therefore,  the  slotted 

17 


18 


SLIDE-VALVES. 


form  of  connection  to  the  valve-rod  were  employed,  it  would 
be  necessary  either  to  place  the  eccentric-pin  at  the  end  of 
the  shaft,  which  is  done  in  some  engines,  or  to  reduce  the 
diameter  of  the  shaft  at  the  point  from  which  the  motion  of 
the  valve  is  derived,  as  shown  in  Fig.  15.  Here  the  shaft 


FIG.  15 

center  is  <9,  and  the  crank-shaft  is  cut  away  to  form  the  pin 
C,  on  the  same  principle  that  center-crank  engines  are  built. 
The  eccentricity  is  OE,  the  center  of  the  pin  being  at  E\ 
and  motion  would  be  imparted  to  the  valve  by  a  rod  attached 
to  C.  To  avoid  these  difficulties,  and  also  to  obtain 
adjustability,  the  connection  shown  in  Fig.  16  is  employed. 


FIG.  16 


Here  the  eccentric-pin  is  expanded  until  it  surrounds  the 
shaft.  This  is  done  by  taking  a  disk  of  greater  diameter  than 
the  shaft,  as  in  Fig.  17,  where  the  center  of  the  disk  is  E.  and 
boring  in  it  a  hole  the  diameter  of  which  is  equal  to  that  of 


CRANKS,   CONNECTING-RODS,  AND    ECCENTRICS.          19 

the  crank-shaft.  The  hole  and  the  disk  are  not  concentric, 
the  center  of  the  former  being  at  O.  The  distance  OE  is  the 
eccentricity,  or  throw  of  the  eccentric. 
The  disk  may  then  be  keyed  to  the  shaft 
in  any  desired  position,  thus  obtaining 
the  power  to  adjust  the  angular  advance. 

Returning  to  Fig.  16,  B  is  the  disk,  now 
called  the  eccentric-sheave,  and  OE  is  the 
eccentricity.  A  ring,  T,  called  the  eccen- 
tric-strap, surrounds  the  eccentric-sheave, 
and  is  free  to  turn  on  it.  The  strap  is  secured  to  the  eccen- 
tric-rod, Nt  and  the  latter  communicates  the  motion  of  the 
eccentric  to  the  valve,  either  by  a  rocker,  or  by  direct  attach- 
ment to  the  valve-rod.  The  motion  of  the  valve  thus  secured 
is  exactly  the  same  as  if  the  eccentric-rod  were  attached  to  an 
eccentric-pin  at  E,  as  shown  by  the  center-lines. 

The  use  of  the  crank  and  connecting-rod  introduces  am 
irregularity  due  to  the  "angularity  of  the  rod."  With  the 
slotted-cross-head  form  of  connection,  it  will  be  remembered, 
the  displacement  of  the  piston  from  the  end  of  its  stroke  is 
exactly  equal  to  the  horizontal  displacement  of  the  crank-pin; 
and,  therefore,  when  the  crank-pin  is  in  its  middle  position 
the  piston  is  at  the  center  of  its  stroke;  and  this  is  true  on 
both  the  forward  and  return  strokes.  The  velocity  of  the 
crank-pin  is  assumed  to  be  uniform,  and  hence  the  piston 
always  takes  the  same  amount  of  time  to  complete  a  half- 
stroke.  Both  strokes  being  alike,  it  follows  that  for  a  given 
piston  displacement  from  the  end  of  the  stroke,  the  angle 
made  by  the  crank-line  with  the  line  of  stroke  is  the  same  at 
both  ends.  This  is  shown  in  Fig.  18,  in  which  XZ  is  the  line 
of  stroke,  AB  is  the  length  of  the  stroke,  and  the  circle  whose 
center  is  O,  and  whose  diameter  MN  is  equal  to  AB,  is  the 
path  of  the  crank-pin.  Now  if  BP  is  the  piston  displacement 
on  the  forward  stroke,  the  crank-pin  position,  C,  correspond- 
ing to  this  is  found  by  taking  PD  equal  to  BN,  and  drawing 


20 


SLIDE-VALVES. 


CD  perpendicular  to  the  line  of  stroke  until  it  cuts  the  circle 
at  C.  Then  is  CO  the  crank  position,  and  CON  is  the  crank 
angle  corresponding  to  the  given  piston  displacement  on  the 
forward  stroke.  Now  let  AP',  equal  to  BP,  be  the  piston 
displacement  on  the  return  stroke.  Take  P'Df,  equal  to  AM, 


FIG. 18 


or,  which  is  the  same,  BN  or  PD,  and  locate  C'  by  drawing 
D ' C'  perpendicular  to  the  line  of  stroke  until  it  cuts  the  circle 
at  C'.  Then  is  C'O  the  crank  position  and  C'OM  is  the 
crank  angle  corresponding  to  the  given  piston  displacement. 
As  MD'  is  equal  to  ND,  both  being  equal  to  the  piston  dis- 
placement, the  angles  C'OM  and  CON  are  equal. 

With  the  crank  and  connecting-rod  arrangement  none  of 
these  things  are  true.  This  is  shown  in  Figs.  19  and  20,  which 
are  lettered  similarly  to  Fig.  18.  The  length  of  the  connect- 
ing-rod is  BN  With  the  piston  in  its  middle  position  on  the 
forward  stroke,  as  at  P,  Fig.  19,  the  crank- pin  position  is  C, 
found  by  taking  P  as  a  center  and  striking  an  arc,  with  the 
length  of  the  rod  as  a  radius,  until  it  cuts  the  circle  at  C. 
Then  is  CO  the  crank  position  for  mid-stroke,  forward,  of  the 
piston,  and  CON  is  the  crank  angle.  During  the  first  half  of 
the  piston  stroke  the  crank-pin  travels  from  N  to  C\  and 


CRANKS,   CONNECTING-RODS,  AND   ECCENTRICS. 


21 


during  the  last  half  the  crank-pin  moves  from  C  to  M.  With 
a  uniform  velocity  of  the  crank-pin,  which  is  assumed,  the 
piston  travels  slower  during  the  last  half  of  the  forward  stroke 


FIG.  19 


than  during  the  first  half.  On  the  return  stroke,  with  the 
piston  at  P,  the  crank  position  is  C'O,  and  the  crank  angle  is 
C'OM,  which  is  evidently  greater  than  CON-  It  is  also 


FIG.  20 


c' 


evident  that  the  piston  velocity  is  less  during  the  first  half  of 
the  return  stroke  than  during  the  last  half;  for  the  crank-pin, 
moving  at  a  uniform  rate,  takes  longer  to  pass  over  MC'  than 
it  does  to  cover  C'N. 


22  SLIDE-  VA L  VES. 

In  Fig.  20  the  piston-displacements  are  the  same  as  in 
Fig.  1 8,  but  the  crank-pin  positions  are  found  by  striking 
arcs  about  P  and  P'  as  centers,  the  radius  in  each  case  being 
the  length  of  the  connecting-rod.  The  crank  angles  C'OM 
and  CON  are  evidently  unequal,  C'OM  being  the  greater. 

The  foregoing  may  be  summed  up  as  follows: 

With  the  ordinary  crank  and  connecting-rod  the  crank 
angles  corresponding  to  equal  piston  displacements  on  the  for- 
ward and  return  strokes  are  unequal,  that  on  the  return  stroke 
being  the  greater. 

With  the  ordinary  crank  and  connecting-rod,  assuming  tJie 
velocity  of  the  crank-pin  to  be  uniform,  the  piston  velocity  is 
not  uniform,  being  less  during  the  crank- end  half  of  each  stroke 
than  during  the  head-end  half. 

The  variation  between  the  crank  angles  on  the  forward  and 
return  strokes  depends  on  the  length  of  the  connecting-rod  as 
compared  with  that  of  the  crank;  the  longer  the  connecting- 
rod,  the  less  will  be  the  difference  between  the  crank  angles 
for  equal  piston  displacements  on  the  forward  and  return 
strokes.  When  the  rod  is  infinitely  longer  than  the  crank, 
there  is  no  difference  between  the  angles;  and  hence  the 
slotted-cross-head  form  of  connection  is  usually  referred  to  as 
the  " infinite  connecting-rod."  With  the  ordinary  types  of 
engines,  where  the  rod  is  from  four  to  ten  times  the  length 
of  the  crank,  the  difference  must  be  considered,  and  affects 
the  amounts  of  lap  required  on  the  ends  of  the  valve  to  secure 
cut-off  at  the  same  point  in  each  stroke,  making  the  laps 
unequal.  This  is  because  the  angle  between  the  crank  and 
eccentric  is  fixed,  and  the  valve  displacement  varies  with  the 
<crank-angle;  hence  if  the  crank-angles  are  unequal,  the  valve 
displacements  are  unequal;  and  the  lap  required  to  cut  off  at 
any  point  in  the  stroke  being  equal  to  the  valve  displacement 
at  that  point,  it  follows  that  unequal  crank-angles  require 
unequal  laps. 

With  the  eccentric  the  case  is  different,  as  the  eccentricity 


CRANKS,   CONNECTING-RODS,  AND    ECCENTRICS.         2$ 

is  so  much  less  than  the  length  of  the  eccentric-rod  that  the 
effect  of  angularity  may  be  neglected.  This  is  because  the 
eccentricity  is  much  less  than  the  throw  of  the  crank,  while 
the  eccentric-rod  is  equal  to  or  longer  than  the  connecting- 
rod;  hence  the  ratio  of  the  eccentric-rod  to  the  eccentricity 
is  much  greater  than  the  ratio  of  the  connecting-rod  to  the 
crank.  The  valve  displacement  is  therefore  reckoned  as  equal 
to  the  eccentric  displacement. 

It  is,  therefore,  necessary  to  determine  the  crank  positions 
corresponding  to  the  piston  positions  at  which  cut-off  is 
desired,  when  designing  a  valve  for  any  engine;  or,  when 
studying  the  action  of  a  valve  already  in  use,  to  determine  the 
piston  position  corresponding  to  a  given  crank  position.  The 
method  of  Figs.  19  and  20  may  be  employed,  but  it  has  the 
disadvantage  of  requiring  to  be  drawn  on  a  large  scale,  and 
of  having  the  lines  intersect  at  acute  angles,  making  it  hard 
to  obtain  good  results.  A  better  method  is  shown  in  Fig.  21. 
This  method  is  a  modification  of  the  original  method  of 
M.  Marcel  Deprez,  and  the  author  acknowledges  his  indebted- 
ness to  the  Practical  Engineer. 

Lay  off  the  straight  line  X  Y  of  indefinite  length,  and  on 
it  take  any  convenient  point,  as  O,  for  a  center;  and  about 
O  describe  a  circle  with  a  radius  equal  to  the  length  of  the 
crank.  This  gives  the  crank-circle,  whose  diameter  is  MN. 
Lay  off  the  distance  OO'  ,  equal  to  the  square  of  the  crank 
length,  divided  by  four  times  the  length  of  the  rod,  or 


where  R  =  length  of  connecting-rod; 
C  =  length  of  crank. 

About  O'  as  a  center,  and  with  a  radius  equal  to  the  length 
of  the  crank,  describe  an  auxiliary  crank-circle,  whose 
diameter  is  M'N'.  The  point  Of  should  always  be  between 


24  SLIDE-  VA  L  VES. 

the  center  of  the  crank-shaft,  O,  and  the  cylinder;  that  is,  in 
the  case  shown,  the  cylinder  is  at  the  right. 

Now  determine  the  crank-pin  position  corresponding  to 
the  given  piston  position  on  the  auxiliary  crank-circle,  as  if 
the  slotted  cross-head  were  used.  For  example,  if  the  piston 
is  at  \  stroke,  take  N'D  equal  to  i  of  N'M',  and  draw  DC 


FIG.  21 

perpendicular  to  XY  until  it  cuts  the  auxiliary  circle  at  c. 
Then  c  is  the  crank-pin  position  on  the  auxiliary  circle. 

Next,  from  c,  draw  a  line  to  O.  Then  the  part  of  this 
line  cO  included  within  the  crank  circle  is  the  crank  position 
corresponding  to  the  given  piston  position;  that  is,  CO  is  the 
required  crank  position.  This  diagram  shows  very  clearly  the 


CRANKS,  CONNECTING-RODS,  AND    ECCENTRICS.         2$ 

difference  between  the  finite  and  infinite  rods.  For  cO'  is 
the  crank  position  corresponding  to  the  given  piston  position 
with  the  infinite  rod,  and  the  crank  angle  cO ' N'  is  evidently 
different  from  CON. 

For  the  return  stroke  the  process  is  similar.  Take  M'D' 
equal  to  the  given  piston  displacement.  Draw  D1 '  c' ',  per- 
pendicular to  the  line  of  stroke  until  it  cuts  the  auxiliary 
crank-circle  at  c1 '.  Join  c'  and  (9,  and  produce  Oc'  until  it 
strikes  the  crank-circle  at  C' .  Then  is  OC1  the  crank  position 
corresponding  to  the  given  piston  displacement. 

A  reversal  of  this  method  will  give  the  piston  displace- 
ment corresponding  to  any  given  crank  position.  Draw  the 
given  crank  position,  and  determine  its  intersection  with  the 
auxiliary  circle,  producing  the  given  crank-line,  if  necessary. 
From  this  intersection  drop  a  perpendicular  upon  the  line  of 
stroke.  The  distance  from  the  foot  of  this  perpendicular  to 
the  dead-center  of  the  auxiliary  crank-circle  gives  the  piston 
displacement. 

It  will  be  seen  at  once  that  the  distance  OO'  varies  with 
the  ratio  of  the  connecting-rod  to  the  crank,  and  that  the 
difference  between  the  crank  angles  depends  on  the  length  of 
00'. 

To  save  calculation,  Table  I  has  been  prepared,  which 
gives  the  values  of  OO'  in  terms  of  the  crank  length.  For 
example,  if  the  stroke  length  is  36  inches  and  the  connecting- 
rod  is  5  feet  3  inches  long,  the  distance  OO'  is  i-g^-  inches,  if 
laid  out  full-size,  found  as  follows:  The  crank  is  one  half  of 
the  stroke,  or  18  inches,  and,  the  connecting-rod  being  5  feet 
3  inches  or  63  inches  length,  the  ratio  of  the  connecting-rod 
to  the  crank  is  63  ^-  18  =  3^.  From  Table  I  it  is  found 
that  for  this  ratio  the  distance  OO  is  equal  to  the  crank 
length  divided  by  14  or  multiplied  by  .0714.  Using  the 
latter  gives  18  X  .0714  =  1.2852  inches.  From  Table  II 
it  is  found  that  the  nearest  fraction  of  an  inch  to  .2852  is 


26  SLIDE-  VA L  VES. 

.28125,   corresponding  to  -/$  of  an  inch.     OO'  is  therefore 
made  i¥9^-  inches. 

If  the  diagram  is  to  be  made  on  any  pther  scale,  OO'  is 
reduced  or  enlarged  in  proportion.  At  half-size  it  would  be 
.6426  or  41.. 


CHAPTER    III. 
VALVE  DIAGRAMS-GENERAL  PRINCIPLES. 

IN  designing  a  new  valve,  or  in  studying  the  action  of  an 
old  one,  it  is  necessary  to  know  the  valve  displacement  corre- 
sponding to  a  given  crank  position.  This  may  be  found  by 
making  a  center-line  sketch  of  the  gear  employed,  but  this 
requires  too  much  space,  is  not  sufficiently  accurate,  and  is 
too  long  and  tedious. 

There  are  various  diagrams  designed  to  show  at  a  glance 
the  relation  between  the  crank  position  and  the  valve  dis- 
placement, all  having  certain  good  points.  The  one  which 
will  be  used  in  this  work  is  that  developed  by  Dr.  Gustave 
Zeuner. 

CASE  I. 

That  of  a  simple  valve,  with  neither  outside  nor  inside  lap; 
angle  between  crank  and  eccentric  90°.  Such  a  valve  is 
shown  in  Figs.  3  to  8. 

Draw  the  two  lines  JTFand  VZ,  Fig.  22,  at  right  angles, 
intersecting  at  O.  This  point  O  represents  the  center  of  the 
crank-shaft,  and  VZ  represents  the  line  of  stroke.  About 
O  as  a  center,  and  with  a  radius  equal  to  the  length  of  the 
crank,  describe  the  crank-circle  as  shown.  Take  OE  on  X Y 
equal  to  one-half  the  eccentricity,  and  describe  the  valve-circle 
shown.  Locate  E'  in  the  same  way,  and  draw  the  lower 
valve  circle. 

Now,  having  this  figure,  the  valve  displacement  for  any 
crank  position  is  equal  to  that  portion  of  the  crank  line  which 

27 


28 


SLIDE-VALVES. 


is  included  in  the  valve-circle.      For  example,  when  the  crank 
is  in  the  position  0i,  the  valve  displacement  is  Oj. 

This  diagram  shows  the  action  of  the  valve  to  be  just  as 
described  in  Chapter  I.     Suppose  the  engine  to  be  running 


over,  as  shown  by  the  arrow;  then  the  action  of  the  valve  OR 
the  head-end  is  as  follows:  With  the  crank  at  A,  the  crank 
line  is  A O,  and  the  valve  displacement  is  zero,  as  AO  does 
not  cut  either  valve-circle.  When  the  crank  moves  ahead  to 
I,  the  valve  displacement  is  Oj.  When  the  crank  reaches  2» 


VALVE  DIAGRAMS— GENERAL   PRINCIPLES.  29 

at  right  angles  to  line  of  stroke,  the  valve  is  displaced  the 
distance  Ov,  which  is  the  eccentricity,  or  half  travel.  While 
the  crank  travels  from  A  to  2  the  port  is  gradually  opening. 
From  2  to  R  the  port  gradually  closes,  until  at  R  it  is  closed. 
The  valve  now  commences  to  reverse  its  direction,  as  shown 
by  the  fact  that  the  crank  line,  as  it  moves  on,  cuts  the  lower 
valve  circle.  The  valve  is  at  the  other  end  of  its  travel  when 
the  crank-pin  is  at  3,  as  shown  by  the  fact  that  the  valve 
displacement  is  Os,  equal  to  the  eccentricity.  When  the 
crank  reaches  OA,  the  valve  is  again  central. 

With  this  type  of  valve  the  port-opening  is  equal  to  the 
valve  displacement.  Therefore,  during  the  entire  forward 
stroke,  or  while  the  crank-pin  moves  from  A  to  R,  the  head- 
end port  is  open  to  admission,  and  during  the  entire  return 
stroke,  or  while  the  crank-pin  goes  from  C  to  A,  the  head-end 
port  is  open  to  exhaust,  as  shown  in  the  figure. 

The  action  of  the  valve  on  the  crank  end  is  directly  the 
reverse;  that  is,  the  valve  displacement  which  opens  the  head- 
end port  closes  the  port  at  the  crank  end.  The  events  of  the 
strokes  are  therefore  read  from-  opposite  valve  circles,  as 
shown  in  Fig.  23.  With  the  crank  at  A'  the  valve  is  central, 
but  as  it  moves  on  the  crank-end  port  is  opened  to  admit 
steam,  by  the  amount  of  the  valve  displacement,  until,  with 
the  crank-pin  at  3,  the  crank-end  port  is  wide  open.  From 
there  on  to  R!  the  port  is  being  closed,  until  at  R  the  valve 
is  central.  Admission  and  exhaust  are  therefore  reversed. 

CASE  II. 

Simple  D-valve,  with  outside  lap;  angle  between  crank 
and  eccentric  90°. 

The  only  difference  between  such  a  valve  as  this  and  that 
of  Case  I  is  that  the  port-opening  for  admission  is  not  equal 
to  the  valve  displacement,  but  is  less  by  the  amount  of  outside 
lap,  as  pointed  out  in  Chapter  I.  The  exhaust  remains 


SLIDE-VALVES. 


unchanged.     The  diagram  for  this  case  is  shown  in  Figs.  24 
and  25. 

In  order  to  find  the  port-opening  for  a  given  crank  posi- 
tion, it  is  only  necessary  to  subtract  the  amount  of  the  lap 


from  the  valve  displacement;  and  this  can  be  done  by  striking 
an  arc  about  O  as  a  center  and  with  a  radius  OL  equal  to  the 
outside  lap.  This  being  done,  the  port-opening  is  given  by 
the  following 

RULE:    The  port- opening  for  any  given  crank  position  is 
equal  to  that  portion    of   the   crank-line   which   is   included 


VALVE   DIAGRAMS— GENERAL   PRINCIPLES.  31 

between  the  lap  and  valve-circles.      For   example,    with    the 
crank-pin  at  2,  Fig.  24,  the  port-opening  is  Lv. 

Fig.  24  represents  the  diagram  for  the  head  end.      Events 
on  the   forward   stroke  are   read   above   the   line  of  stroke; 


events  on  the  return  stroke  are  read  below  it,  the  direction  of 
rotation  of  the  engine  being  shown  by  the  arrow.  With  the 
crank  at  T  there  is  no  displacement  of  the  valve.  When  it 
gets  to  A,  the  displacement  is  just  equal  to  the  lap,  and  any 
further  movement  of  the  crank  will  open  the  head-end  port 


OF  THE 


UNIVERSITY 
<>A 


SLIDE-VALVES. 


to  steam.  Therefore  admission  begins  at  A ;  and  it  lasts 
until  the  port  is  again  closed,  which  will  be  when  the  crank 
is  at  C,  when  the  valve  displacement  is  equal  to  the  outside 
lap.  This  gives  the  two  following  rules: 


"Admission"  begins  when  the  crank  line  passes  through 
the  first  intersection  of  the  valve  and  outside-lap  circles.  OA 
passes  through  Oa. 

"Cut-off"  occurs  when  the  crank  line  passes  through  the 
second  intersection  of  the  valve  and  lap  circles. 


VALVE  DIAGRAMS— GENERAL   PRINCIPLES.  35 

In  Fig.  24  the  lap-circle  is  only  drawn  above  the  line  of 
stroke.  This  is  because  the  outside  lap  does  not  affect  the 
exhaust,  and  the  exhaust  and  admission  are  read  from  opposite 
valve-circles. 

The  exhaust  opens  at  R,  and  closes  at  T — that  is,  it  lasts 
during  the  entire  stroke,  just  as  in  Case  I. 

From  C  to  R  no  steam  is  admitted,  and  the  steam 
admitted  up  to  cut-off  expands. 

From  T  to  A  the  cylinder  is  empty,  and  the  engine  runs 
only  by  virtue  of  the  momentum  of  the  fly-wheel. 

Fig.  25  shows  the  diagram  for  the  crank-end  of  the 
cylinder,  the  valve  being  supposed  to  have  the  same  outside 
lap  on  both  ends. 

CASE  III. 

Simple  D-valve,  both  outside  and  inside  lap;  angle 
between  the  crank  and  eccentric  90°. 

This  case  is  shown  in  Figs.  26  and  27,  the  former  being 
for  the  head-end,  and  the  latter  for  the  crank-end.  The  valve 
is  supposed  to  have  the  same  amount  of  inside  lap  at  both 
ends.  The  valve  travel,  crank  throw,  and  outside  lap  are  the 
same  as  in  the  preceding  problem. 

The  inside  lap  affects  only  the  exhaust;  and  as  its  effect 
is  to  lessen  the  amount  that  the  port  is  open  to  exhaust  by 
the  amount  of  lap,  allowances  can  be  made  for  it  by  striking- 
an  arc  about  O  as  a  center,  and  with  a  radius  OI  equal  to  the 
inside  lap,  in  a  manner  similar  to  that  employed  for  the  out- 
side lap.  This  gives  as  a  rule: 

The  port-opening  to  exhaust  corresponding  to  any  given 
crank  position  is  equal  to  that  portion  of  the  crank-line  which 
is  included  between  the  valve-circle  and  inside  lap-circle.  The 
exhaust-opening  and  steam-opening  are  read  from  different 
valve  circles,  it  must  be  remembered. 

Fig.  26  is  the  same  as  Fig.  24  as  far  as  the  facts  relating 


34 


{SLIDE-VALVES. 


to  admission  and  cut-off  are  concerned.  The  exhaust  does 
not  open  until  the  valve  displacement  is  equal  to  the  inside 
lap,  or  until  the  crank  is  at  R.  It  remains  open  while  the 


cranK  passes  from  R  to  T,  when  the  valve  displacement  is 
again  equal  to  the  inside  lap.  From  which  these  two  rules: 

"Release"  or  exhaust -opening,  occurs  when  the  crank-line 
passes  through  the  first  intersection  of  the  valve  and  inside  lap 
circles.  OR  passes  through  r. 

"  Compression,"  or  exhaust-closure,  occurs  when  the  crank 


VALVE   DIAGRAMS— GENERAL   PRINCIPLES. 


35 


line  passes  through  the  second  intersection  of  the  valve  and  in- 
side-lap circles.      OT  passes  through  /. 

From  C  to  M  expansion   occurs;   and  a  peculiarity  of  this 
type  of  valve  is  that  from  M  to  R  the  steam  is  compressed. 


FIG. 27 

This  is  evident  because  the  piston  reverses  the  direction  of 
its  motion,  and  the  exhaust  does  not  open  until  the  piston 
has  passed  over  a  portion  of  its  return  stroke.  In  the  same 
way,  compression  occurs  from  T  to  N,  and  from  N  to  A  this 
compressed  steam  expands. 


SLIDE-VALVES. 


CASE  IV. 

Simple  D-valve,  with  both  inside  and  outside  laps,  with 
the  angular  advance  sufficient  to  open  the  port  to  steam  at 
the  beginning  of  the  stroke. 

The  diagrams  for  this  case  are  shown  in  Figs.  28  and  29, 


<o- 


ADMISSION 


to' 


FIG.  28 


Fig.  28  being  for  the  head-end.      These  figures,   as  in  the 
previous  cases,  are  lettered  alike,  the  sole  difference  being  that 


VALVE  DIAGRAMS— GENERAL   PRINCIPLES. 


37 


the  letters  are  primed  in  the  crank-end  diagram.  The  nota- 
tion shown  and  used  thus  far  will  be  used  throughout  the 
work. 

In  order  to  make  the  valve-circle  pass  through  the  inter- 
section of  the    outside  lap-circle  and  the  valve-circle,  it  is 


I 

o 
o 


V  I 


ADMISSION 


L'l 


2         ~~^+5 
.V 


I 


Y 

FIG.  29 


necessary  to  incline  the  line  DD'  joining  E  and  /?•',  as  shown 
in  the  cuts.  The  angle  DOX  is  equal  to  the  angular  advance ; 
and  it  makes  no  difference  whether  the  eccentric  leads  or 
follows  the  crank — that  is,  whether  the  angle  between  the 


38  SLID  E-  VA  L  VES. 

crank  and  eccentric  is  90°  +  the  angular  advance,  or  90°  — 
the  angular  advance — the  \mzDD'  is  always  inclined  as  shown, 
with  the  upper  end,  D,  brought  over  toward  the  cylinder. 
The  angular  advance  will  take  care  of  itself  when  it  comes  to 
the  valve-setting.  Nor  does  it  make  any  difference  whether 
the  cylinder  is  to  the  right  or  left  of  the  crank-shaft  as  it  is 
located  in  the  diagram.  The  diagrams  are  for  the  head-end 
and  crank-end,  which  disposes  of  that  question.  Neither 
does  it  make  any  difference  whether  the  engine  runs  over  or 
under.  This  is  one  of  the  beauties  of  the  diagram — that  it 
fits  any  and  all  circumstances. 

The  dimensions  of  the  valve  being  taken  the  same  as  in 
the  preceding  case,  it  will  be  seen  at  once  that  the  diagram 
shows  the  effect  of  angular  advance  to  be,  as  stated  in 
Chapter  I,  that  of  making  all  the  events  of  the  stroke  earlier. 

While  the  crank  is  traveling  from  5  to  4  the  valve  moves 
in  one  direction ;  and  while  the  crank  travels  from  4  to  5  the 
motion  of  the  valve  is  reversed.  This  will  help  to  explain 
why  the  facts  relating  to  exhaust  and  admission  are  not  read 
from  the  same  valve  circles.  For,  referring  to  Figs.  23  to 
28,  it  will  be  seen  that  when  the  valve  leaves  its  central 
position  and  moves  to  the  left,  it  uncovers  the  head-end  port 
to  steam,  and  continues  to  leave  it  open  until  it  is  again 
central  on  its  travel  to  the  right.  That  is,  the  steam-opening 

is  affected  by  the  last  half  of  the  left-hand  travel  of  the  valve 

— ^ 
and  the  first  half  of  the  right-hand  travel.     Then,  in  all  the 

valve  diagrams,  Ov  represents  the  last  half  of  the  travel  in 
one  direction,  and  vO  represents  the  first  half  of  the  travel  in 
the  other  direction.  The  exhaust  on  the  head-end,  Figs.  3 
to  8,  is  affected  in  the  opposite  way,  that  is,  by  the  first  half 
of  the  left-hand  travel  and  the  last  half  of  the  right-hand 
travel,  a-nd  is  therefore  read  from  the  other  valve-circle,  where 
Os  is  travel  in  one  direction,  and  sO  is  travel  in  the  other. 


VALVE  DIAGRAMS— GENERAL   PRINCIPLES. 


39 


CASE  V. 

Simple  D-valve,  both  inside  and  outside  laps,  and  having 
the  angular  advance  sufficient  to  o^pen  the  port  before  the 
beginning  of  the  stroke. 

This  case,  which  is  shown  in   Figs.  30  and   31,  simply 


EIG.  30 

amounts   to    moving   the    eccentric   a    little    farther   ahead. 
Admission  occurs  when  the  crank  is  at  A.     The  amount  de, 


40 


SLIDE-VALVES. 


which  the  port  is  opened  when  the  piston  begins  its  stroke, 
is  called  the  "  steam-lead,"  or  simply  "  lead."  The  angle 
ZOA  is  called  the  "angle  of  lead,"  or  "lead  angle.' 


FIG.  31 

Similarly,  fg  is  the  exhaust-lead,  and  ROM  is  the  exhaust- 
lead  angle. 

The  "lead"  angle  is  the  angle  between  the  crank  line  at 
Admission  and  the  line  of  stroke.  ZOA  is  the  lead  angle. 

This  case  is  the  general  one.  The  instructions  for  drawing 
it  may  be  summed  up  as  follows: 


VALVE  DIAGRAMS— GENERAL   PRINCIPLES.  4! 

Draw  the  two  lines  XY  and  VZ  at  right  angles,  intersect- 
ing at  (9,  which  represents  the  center  of  the  crank-shaft,  VZ 
representing  the  line  of  stroke.  About  O  as  a  center,  and 
with  a  radius  OM  equal  to  the  crank,  describe  the  crank-circle 
whose  diameter  is  MN,  which  is  equal  to  the  stroke  of  the 
engine.  Lay  off  the  angle  XOD  equal  to  the  angular  advance. 
Draw  DOD'.  Lay  off  OE  equal  to  half  of  the  eccentricity, 
and  with  E  as  a  center  and  a  radius  OE  describe  the  upper 
valve-circle.  Locate  E'  and  describe  the  lower  valve-circle 
in  the  same  way.  (It  is  neither  necessary  nor  advisable  to 
use  the  same  scale  for  the  valve  and  crank-circles.  The 
crank-circle  is  only  employed  to  determine  the  inclination  of 
the  crank  to  the  line  of  stroke,  and  need  be  drawn  only  on  a 
small  scale.)  About  O  as  a  center,  and  with  a  radius  OL  equal 
to  the  outside  lap,  describe  the  outside  lap  circle.  About  O 
as  a  center,  and  with  a  radius  Of  equal  to  the  inside  lap, 
describe  the  inside  lap  circle. 

The  diagram  is  then  read  in  accordance  with  the  rules 
given. 


EFFECT   OF   CHANGING  VARIOUS  DIMENSIONS  OF  THE  VALVE, 

AS   SHOWN   BY    THE    DIAGRAM,    WHEN   THERE   IS 

ANGULAR   ADVANCE. 

Angular  Advance. — The  effect  of  changing  this  is  shown 
in  Figs.  28  and  30,  or  29  and  31.  If  the  angular  advance  is 
increased,  its  effect  is  to  make  all  the  events  of  the  stroke 
earlier.  If  it  is  decreased,  all  the  events  occur  sooner. 
Refer  to  Table  III. 

Valve  Travel. — If  this  is  increased,  cut-off  and  compres- 
sion come  later  and  admission  and  release  occur  earlier.  If 
it  is  decreased,  cut-off  and  compression  occur  earlier  and 
admission  and  release  come  later.  Refer  to  Fig.  32  and 
Table  III. 


SLIDE-VALVES. 


Outside  Lap. — Increasing  this  makes  admission  later  and 
cut-off  earlier,  prolongs  the  expansion  and  compression, 
decreases  the  lead,  and  leaves  the  exhaust  unchanged.  De- 
creasing the  outside  lap  increases  the  lead,  makes  cut-off  later 


F 

FIG.  32 

and  admission  earlier,  shortens  expansion  and  compression, 
and  does  not  affect  the  exhaust.  Refer  to  Fig.  33  and 
Table  III. 

Inside  Lap. — Increasing  this  does  not  affect  lead,  admission, 
or  cut-off,  but  it  makes  release  later  and  compression  earlier, 


*& 


L^ 


OF  THB 


UNIVERSITY 
^^  0 ..    ""     ~;     »    ^ 

VA  L  VE  D I  A  GRA  MS— GENERA  L^Sf&ef&fES. 


43 


thus  prolonging  expansion  and  compression.     Decreasing  the 
inside  lap  makes  release  earlier  and  compression  later,  thus 


\o« 
V 


V 


ADMISSION 


FIG.  33 

shortening  both   compression  and   expansion.     See  Fig.   33 
and  Table  III. 


CHAPTER    IV. 
DIMENSIONS  OF  PORTS,  STEAM-PIPES,  AND   BRIDGES. 

A  STUDY  of  the  valve  diagrams  will  show  at  once  that  the 
dimensions  of  the  valve  are  influenced  very  strongly  by  the 
point  of  cut-off.  Next  in  importance  to  this,  as  a  controlling 
influence,  is  the  width  of  the  steam-port.  This  latter  is  not 
assumed  arbitrarily,  but  is  calculated  with  a  reasonable  degree 
of  accuracy. 

When  steam  passes  through  an  opening  at  a  greater  speed 
than  6000  or  8000  feet  per  minute  it  is  choked,  throttled,  or 
(<  wire- drawn  "  ;  that  is,  its  pressure  after  passing  through  the 
opening  is  less  than  the  pressure  urging  it  through.  For  this 
reason  the  steam  must  not  be  forced  to  pass  through  the 
ports  at  a  greater  speed  than  6000  feet  per  minute.  Now,  in 
order  to  fill  the  cylinder  up  to  cut-off  with  steam  at  boiler- 
pressure,  it  is  necessary  that  the  volume  of  steam  admitted 
shall  be  equal  to  the  volume  swept  through  by  the  piston. 

The  volume  of  the  cylinder  up  to  cut-off  is  of  course  equal 
to  the  area  of  the  piston  multiplied  by  the  distance  through 
which  the  piston  has  traveled.  If  the  piston  area  is  one  square 
foot,  and  the  travel  up  to  cut-off  is  2  feet,  then  1X2  =  2 
cubic  feet  of  steam  must  be  admitted  while  the  piston  is 
moving  through  those  two  feet.  The  amount  of  steam  to  be 
supplied  can  also  be  found  by  multiplying  the  area  of  the 
piston  in  square  feet  by  the  piston  speed  in  feet  per  minute, 
and  multiplying  this  result  by  the  time  in  minutes  that  the 
port  is  open.  This  may  be  expressed  in  a  formula  as  follows: 

C=  A  X  SX  T,  .     .     .     .     .     .     (i) 

44 


DIMENSIONS  OF  PORTS,  STEAM-PIPES,  AND  BRIDGES.   45 

where   C  '=  volume  in  cubic  feet  up  to  cut-off; 
A  =  piston  area  in  square  feet; 
5  =  piston  speed  in  feet  per  minute; 
T—  time  in  minutes  that  port  is  open. 

For  example,  if  the  piston  area  is  I  square  foot,  the  pis- 
ton speed  400  feet  per  minute,  and  the  port  is  open  ^^  of  a 
minute,  then 

C  —  i  X  400  X  -   -  =  2  cubic  feet. 
200 

Now,  this  steam  must  be  admitted  through  the  port,  which 
is  open  for  the  same  length  of  time.  The  volume  of  steam 
thus  admitted  is  equal  to  the  velocity  of  the  steam  in  feet  per 
minute  multiplied  by  the  time  the  port  is  open,  and  this 
result  then  multiplied  by  the  area  of  the  port  in  square  feet. 
Expressed  in  a  formula  this  is 

C=  VX  PX  T,  ......     (2) 

where  C  =  volume  admitted  in  cubic  feet; 

V  •=.  velocity  of  steam  in  feet  per  minute; 
T=  time  in  minutes  that  port  is  open. 
For  example,  if  the  velocity  of  the  entering  steam  is  6000 
feet  per  minute,  the  port  area  ^  of  a  square  foot,  and  the 
port  is  open  ^.J^  of  a  minute,  then 


C  =  6000  X  —  X  -  =  2  cubic  feet. 

The  results  given  by  formulas  (i)  and  (2)  must  be  equal, 
of  course,  so  that 

A  X  SX  T=  FXPX  T, 
or,  as  T  is  on  both  sides,  it  may  be  dropped,  giving 

A  X  5=  VXP.  .     .     .     .  ..»-,.     (3) 


46  SLIDE-  VA  L  VES. 

In  designing  a  valve,  the  area  of  the  piston  is  known,  and 
the  velocity  of  the  steam  is  assumed  in  accordance  with  the 
statement  made  above.  The  piston  speed  is  found  by  multi- 
plying the  length  of  the  stroke  by  twice  the  number  of 
revolutions  per  minute;  because  the  piston  makes  two  strokes 
for  every  revolution.  That  is, 


.......      (4) 

where  N  =•  number  of  revolutions  per  minute; 
L  —  length  of  stroke  in  feet. 

Now,  from  formula  (3), 


or 

piston  area  X  piston  speed 
port  area  —  --  ;  —  r-  ---  -f  —          -  . 
velocity  of  steam 

This  may  be  expressed  in  the  following: 

RULE. 

To  Find  the  Port  Area  for  a  Given  Engine  :  Multiply  the 
piston  area  in  square  feet  by  the  piston  speed  in  feet  per  minute  ', 
and  divide  the  product  by  the  velocity  of  the  steam,  in  feet  per 
minute.  The  result  will  be  the  area  of  the  port  in  square  feet. 

For  example,  take  an  engine  18  X  24  —  that  is,  one  in 
which  the  cylinder  is  18  inches  in  diameter  and  24  inches 
stroke,  —  running  at  200  revolutions  per  minute,  and  assume 
the  velocity  of  the  steam  to  be  6000  feet  per  minute.  To  find 
the  port  area  required.  The  piston  speed  is,  according  to 
formula  (4), 

24 
2  X  —  r  X  ioo  =  400  feet. 

The  24  is  divided  by  12  to  reduce  the  stroke  to  feet.  The 
area  of  an  1  8-inch  circle  is  254.47  square  inches,  from  Table  V 
at  the  end  of  the  book,  or  254.47  -f-  144  —  1.766  square  feet. 


DIMENSIONS  OF  PORTS,  STEAM-PIPES,  AND  BRIDGES.   47 

Now,  applying  the  rule,  1.766  X  400  =  706.4,  and  706.4  -f- 
6000  =  .1174  square  feet,  or  16.91  square  inches. 

The  work  of  this  is  renderedve  ry  light  by  using  Table  IV, 
which  contains  the  values  of 

piston  speed 


velocity  of  steam 

for  various  velocities  of  steam,  and  for  piston  speeds  from  100 
to  1200  feet  per  minute,  advancing  by  25.  It  is  only  neces- 
sary to  multiply  the  figure  given  in  the  table  by  the  piston 
area  to  get  the  port  area.  Take  the  example  just  worked. 
Under  6000,  in  the  column  headed  "  Port  Area,  Piston  Area 
as  Unity,"  and  opposite  400  feet  piston  speed,  will  be  found 
.067.  Multiplying  the  area  of  piston,  254.47  square  inches, 
by  .067,  gives  17.049  square  inches.  The  difference  between 
this  result  and  the  one  obtained  by  the  longer  method  is  very 
slight,  only  .13  of  a  square  inch,  and  is  due  to  the  use  of 
decimals. 

The  length  of  the  port  should  be  made,  as  nearly  as  possi- 
ble, equal  to  the  diameter  of  the  cylinder,  and  the  width  is 
of  course  found  by  dividing  the  area  found  as  above  by  the 
length.  If  the  length  of  the  port  is  nine  tenths  of  the 
diameter  of  the  cylinder,  the  width  may  be  found  by  multi- 
plying the  diameter  of  the  piston  by  the  figure  given  in 
Table  i  V.  For  example,  in  the  engine  just  considered,  if  the 
port  length  is  .9  of  the  cylinder  diameter,  the  port  width  is 
found  by  mutiplying  18  (the  diameter  of  the  cylinder)  by 
.058,  which  is  the  figure  opposite  the  piston  speed,  400,  and 
under  "  6000,  Width  of  Port."  This  gives  18  X  .058  = 
1.024  inches  as  the  width  of  the  port. 

If  the  velocity  of  the  steam  were  assumed  as  4000,  the 
multiplier  would  be  .086.  If  the  velocity  were  8000  and  the 
piston  speed  300,  the  multiplier  would  be  .033. 

After  the  steam  has  been  expanded  in  the  cylinder  it  is  at 
a  lower  pressure  than  when  it  was  admitted,  and  will  travel 


4  SLID  E-  VA  L  VES. 

at  a  lower  rate  of  speed.  The  exhaust-ports  should  therefore 
be  made  with  a  greater  area  than  the  steam-ports.  In  order 
to  allow  for  this  it  is  usual  to  assume  that  the  velocity  at 
entrance  is  6000  feet  per  minute,  and  at  exhaust  is  4000  feet 
per  minute.  These  velocities  are  exceeded  in  certain  cases, 
such  as  locomotives,  and  the  table  has  been  extended  to  cover 
these  excesses;  but  throughout  this  work  the  figures  just 
mentioned  will  be  employed. 

The  area  of  the  exhaust-ports  may  be  found  by  applying 
the  rule  given,  or  the  table  may  be  employed. 

Of  course  when  the  same  port  is  used  for  admission  and 
exhaust,  as  in  the  case  of  the  common  slide-valve,  the  port 
width  must  be  made  that  corresponding  to  the  lower  velocity, 
in  order  that  the  exhaust  may  be  free  and  unimpeded. 

The  diameter  of  the  steam-  and  exhaust-pipes  for  a  given 
engine  may  be  determined  by  the  following  rule: 

To  Find  Diameter  of  Steam-  and  Exhaust-pipes  :  Multiply 
the  diameter  of  the  piston  by  the  square  root  of  the  port 
area;  the  latter  being  expressed  as  a  fraction  of  the  piston 
area. 

This  rule  is  obtained  as  follows: 

The  piston  area  is  taken  as  unity.  As  explained  before, 
the  port  area  is  a  certain  fraction  of  the  piston  area.  Repre- 
sent this  fraction  by  F.  The  areas  of  two  circles  are  to  each 
other  as  the  squares  of  their  diameters,  so  that 

I  :  F  : :  D*  :  d\ 

where  I  =  piston  area; 

F=  port  area,  as  a  fraction  of  piston  area,   given  in 

column  of  table  headed  "  Port  Area"; 
D  =  diameter  of  piston; 
d  =  diameter  of  pipe. 
This  formula  gives 


DIMENSIONS  OF  PORTS,  STEAM-PIPES,  AND  BRIDGES.   49 

In  order  to  save  the  trouble  of  the  extraction  of  the  square 
root,  a  column  in  Table  IV,  headed  "  Diameter  of  Steam- 
Pipe,"  has  been  provided,  which  contains  the  values  of  VF 
corresponding  to  different  piston  speeds.  This  column  is  used 
in  the  same  manner  as  the  others.  For  instance,  for  the 
engine  just  discussed,  18  inches  diameter  and  400  feet  piston- 
speed,  the  steam-pipe  diameter  is  18  X  .258  =  4.82  inches; 
the  figure  .258  being  under  "  6000,  Diameter  of  Steam-Pipe  " 
and  opposite  400.  The  exhaust-pipe  diameter  is  found  from 
he  table  by  assuming  the  lower  velocity.  In  this  case  it 
would  be  18  X  .316  =  5.688  inches. 

Suppose  an  engine  designed  for  a  slide-valve  having  a  port 
width  E,  Fig.  34,  equal  to  that  required  for  free  exhaust. 
If  the  port  is  opened  its  full  width,  the  valve  assumes  the 
position  shown  in  the  figure  when  at  the  end  of  its  travel.  It 
is  only  necessary,  however,  to  open  the  steam-port  sufficiently 
to  allow  a  free  admission;  that  is,  the  valve  assumes  the  posi- 
tion shown  in  Fig.  35,  A  being  the  width  required  for 
unchoked  admission.  By  designing  a  valve  to  open  the  port 
only  the  smaller  distance,  the  travel  can  be  decreased  by  the 
difference  between  E  and  A.  This  will  not  interfere  with  a 
free  opening  for  exhaust,  as  this  event  is  controlled  by  the 
inner  edge  of  the  valve;  and  the  lap  there  can  be  so  modified 
as  to  give  the  desired  opening.  As  shown  in  Fig.  35,  the 
exhaust  is  fully  open. 

When  the  valve  has  reached  its  extreme  position,  one  port 
is  open  to  its  greatest  extent  to  admit  steam,  and  the  other 
one  is  opened  as  fully  as  possible  to  exhaust.  The  amount 
that  the  port  is  open  to  steam  in  this  position  is  called  the 
"  maximum  port-opening." 

The  maximum  port-opening \§  made  equal  to  the  width  of 
port  necessary  for  free  admission  of  steam. 

The  maximum  opening  for  exhaust  is  equal  to  the  width  of 
the  port. 

When  the  travel  of  the  valve  is  so  great  that  the  valve 


5°  SLIDE-VALVES. 

assumes  the  position  shown  in  Fig.  36  when  in  its  extreme 
position,  the  amount  x  that  it  runs  over  the  edge  is  called  the 
overtravel. 


In  order  to  fix  these  principles,  as  well  as  to  illustrate  the 
use  of  the  tables  again,  another  example  will  be  worked  out. 

GIVEN.  An  engine  20  X  30,  running  at  150  revolutions. 
Length  of  ports,  19  inches. 

REQUIRED,  (i)  Width  of  port.  (2)  Diameter  of  steam- 
pipe.  (3)  Diameter  of  exhaust-pipe.  (4)  Maximum  port- 
opening. 

150  X  20  X  2 
SOLUTION,   (i)  Piston  speed =  500  feet  per 

minute.  See  formula  (4).  In  Table  I,  under  "4000,  Area 
of  Port"  and  opposite  500,  is  .125.  The  area  of  a  2O-inch 
circle  is  314. 16  square  inches.  314.16  X  .125  =  39.27  square 


DIMENSIONS  OF  PORTS,  STEAM-PIPES,  AND  BRIDGES.    5 


inches.        39.27  -f-  19  =  2.668.        Nearest    sixteenth, 
Width  of  port  =  2\\. 

(2)  In  Table  I,  under  "  6000,  Diameter  of  Steam-Pipe  " 
and  opposite  500,  is  .288.      Then  the  diameter  of  steam-pipe 
should  be  20  X  .288  =  5.76  or  $£  inches. 

(3)  In  Table  I,  under  "4000,    Diameter  of  Steam-Pipe" 


FIG.  36 

and  opposite  500,  is  .353.     Then  the  diameter  of  the  exhaust- 
pipe  should  be  20  X  -353  —  7.06  or  7TV  inches. 

(4)  The  velocity  of  the  steam  at  exhaust  being  4000,  while 
that  at  exhaust  is  6000,  it  follows  that  the  port  need  only  be 
opened  |£§£  of  its  width  to  secure  free  admission.  This  is 
equal  to  f.  The  maximum  port-opening  is  therefore  f  X  2|£ 
=  1.79  inches.  It  is  therefore  made  iff-  inches. 

The  same  result  would  be  obtained  by  using  the  table. 

That  part  of  the  cylinder-casting  which  divides  the  steam- 
and  exhaust-ports  is  called  the  "  bridge."  B,  Fig.  36,  is  the 
width  of  the  bridge.  This  width  must  be  sufficient  so  that  if 
the  valve  has  overtravel — that  is,  if  the  end  of  the  valve  runs 
beyond  the  edge  of  the  port,  as  shown  in  the  figure — the 
steam-  and  exhaust-ports  will  not  be  in  communication.  If 
the  greatest  opening  of  the  port  does  not  exceed  the  width 
of  the  port,  it  is  sufficient  to  make  the  bridge  width  equal  to 
the  thickness  of  the  cylinder-casting,  which  makes  it  compara- 
tively easy  to  obtain  a  good  casting;  but  if  the  valve  has 


5  2  SLIDE-  VA  L  VES. 

overtravel  greater  than  the  thickness  of  the  cylinder-walls,  it 
is  necessary  to  increase  the  thickness  of  the  bridge ;  which 
may  be  summed  up  in  the  following  rule: 

The  Width  of  tJie  Bridge  is  made  equal  to  the  thickness  of 
the  cylinder-wall  when  the  valve  has  either  no  overtravel  or 
an  overtravel  less  than  the  thickness  of  the  cylinder-walls. 
When  the  overtravel  exceeds  this  amount,  the  bridge  thick- 
ness is  found  by  adding  J  inch  to  the  port-opening  and  sub- 
tracting the  width  of  the  port  from  the  sum.  The  remainder 
is  the  thickness  of  the  bridge. 

Fig.  36  shows  the  valve  in  its  extreme  left-hand  position. 
The  exhaust  is  then  passing  through  the  port  P  and  the  open- 
ing O.  These  two  openings  must  be  equal  in  order  that  the 
exhaust  shall  not  be  cramped;  for  it  was  determined  that  the 
width  P  is  necessary  for  free  exhaust  from  the  cylinder,  and 
the  same  width  is  evidently  necessary  at  the  exhaust,  in 
order  not  to  increase  the  velocity  of  steam  at  exhaust  to  such 
an  extent  as  to  increase  the  back  pressure.  The  valve  shown 
in  the  figure  has  no  inside  lap;  and  as  the  valve  has  moved  a 
distance  E,  equal  to  one  half  the  valve  travel  from  its  middle 
position,  the  distance  D,  which  is  the  width  of  the  exhaust- 
port,  must  be  equal  to  E  +  O  —  B\  but  O  is  equal  to  P,  so 
that 

D  =  E  +  P  -  B. 

If  the  valve  has  inside  lap,  the  width  of  the  exhaust-port 
must  be  increased  by  the  amount  of  the  inside  lap. 

To  Determine  the  Width  of  the  Exhaust-port :  i .  When 
the  valve  has  no  inside  lap,  add  together  the  width  of  the 
steam-port  and  one  half  the  valve  travel.  From  the  sum  sub- 
tract the  width  of  the  bridge.  The  remainder  is  the  width  of 
the  exhaust-port.  D  =  E  -f-  P  —  B. 

2 .   When  the  valve  has  inside  lap,  add  together  the  width  of 
the  steam-port  and  the  eccentricity  or  half-travel  of  the  valve. 
From  the  sum  thus  obtained  subtract  the  width  of  the  bridge, 


DIMENSIONS  OF  PORTS,  STEAM-PIPES,  AND  BRIDGES.    53 

and  to  the  remainder  add  the  amount  of  the  inside  lap.      The 
sum  is  the  width  of  the  exhaust-port.     Z>  —  E  -{-  P  —  B  —  I. 

3.  When  the  inside  lap  is  negative — that  is,  when  the  valve 
allows  the  port  to  be  open  to  exhaust  when  the  valve  is  in  its 
middle  position — add  together  the  half -travel  or  eccentricity  and 
the  width  of  the  steam-port.  Then  add  together  the  amount  of 
the  negative  lap  or  inside  clearance  and  the  thickness  of  the 
bridge.  Subtract  this  latter  sum  from  the  first,  and  the  re- 
mainder is  the  zvidth  of  the  exhaust-port.  D—  E  -\-  P—  B  —  /. 


CHAPTER   V. 
VALVE   DIAGRAMS— GENERAL   PROBLEMS  OF   DESIGN. 

THERE  are  several  different  problems  which  may  present 
themselves  for  solution,  circumstances  beyond  the  designer's 
control  having  fixed  certain  dimensions.  But  the  following 
problems  will  cover  the  probable  cases  very  completely. 

PROBLEM  I. 

GIVEN.  Eccentricity,  point  of  cut-off,  angle  of  advance, 
and  point  of  compression. 

REQUIRED.  Lap,  exhaust-lap,  lead,  exhaust-lead,  and 
greatest  possible  openings  of  port  to  both  admission  and 
exhaust. 

SOLUTION.  The  solution  of  this  problem  is  shown  in  Fig. 
37.  Draw  XY  and  VZ  at  right  angles,  intersecting  at  O. 
Draw  DODf,  making  the  angle  DOX  equal  to  the  given  angle 
of  advance.  Find  CO,  the  crank  position  corresponding  to 
the  given  point  of  cut-off  by  either  of  the  two  methods  pre- 
viously described.  It  is  best  to  make  the  determinations  of 
crank  positions  on  a  separate  sheet,  and  transfer  them  to  the 
sheet  on  which  the  solution  is  being  made.  Then  lay  off  ov 
equal  to  the  eccentricity  or  half-travel.  Find  E,  the  middle 
point  of  ov,  and  about  E  as  a  center,  and  with  a  radius  equal 
to  EO,  describe  the  upper  valve  circle.  In  a  similar  way  find 
R'  and  draw  the  lower  valve  circle. 

Now  the  upper  valve  circle  intersects  oc  at  c.  Then, 
according  to  Chapter  III,  oc  must  be  the  outside  lap.  Then 

54 


VALVE-DIAGRAMS—GENERAL   PROBLEMS   OF  DESIGN.    55 


draw  in  the  lap  circle  about  O  as  a  center  and  with  a  radius 
equal  to  oc.  This  will  cut  VZ  at  d,  and  the  distance  de"  from 
d  to  e>  the  intersection  L£  the  valve  circle  with  VZ,  is  the 
lead.  The  lap  circle  cuts  the  valve  circle  at  a,  and  therefore 


OA,   drawn   through    the    point  a  is    the    crank-position    at 
admission.      The  angle  ZOA  is  the  lead  angle. 

Now  draw  in  OT,  the  crank-position  corresponding  to 
the  given  point  of  compression.  This  intersects  the  lower 
valve  circle  at  /,  and  Ot  is  therefore  the  amount  of  the  inside 
lap.  The  inside  lap-circle  is  then  drawn  in  about  O  as  a 
center  and  with  Ot  as  a  radius.  Its  intersection  with  the 


5&  SLIDE-  VA  L  VES. 

valve-circle  at  r  determines  OR,  the  crank-line  at    release. 
The  exhaust  lead  isj^,  and  the  exhaust  lead  angle  is  VOR. 


PROBLEM  II. 

GIVEN.  The  lap,  point  of  cut-off,  and  lead. 
REQUIRED.  The  valve  travel  and  angle  of  advance. 
SOLUTION.   Fig.  38  shows  one  solution  of  this  problem. 
The  lines  XY  and  VZ  are  drawn  at  right  angles,  intersecting 


FIG.  38 


at  O.  The  crank-position  at  cut-off  is  OC,  as  given.  Then 
the  lap-circle  is  drawn  in  about  O  as  a  center  and  with  a 
radius  equal  to  the  given  lap.  Its  intersection  c  with  OC  is 
a  point  through  which  the  valve-circle  must  pass.  O  is 
another  point  on  the  valve-circle.  Another  point  is  ^,  found 
by  laying  off  de  equal  to  the  given  lead.  These  three  points 


VALVE-DIAGRAMS—GENERAL   PROBLEMS   OF  DESIGN.    57 

serve  to  determine  the  center  of  the  valve-circle,  as  follows: 
Find  m,  the  middle  point  of  OC,  and  draw  my  perpendicular 
to  Oc.  Find  n,  the  middle  point  of  Oe,  and  draw  nx  perpen- 
dicular to  Oe.  Then  nx  and  my  intersect  at  E,  which  is  the 
center  required.  Then  draw  in  the  valve-circle  about  E  as  a 
center  and  with  EO  as  a  radius.  Draw  DEO,  and  its  intersec- 
tion v  with  the  valve-circle  determines  Ov,  which  is  the 
eccentricity.  The  valve-travel  is  twice  the  eccentricity. 

Another  solution  is  shown  in  Fig.  39.      Having  found  the 
points  c  and  ^,  draw  ev  perpendicular  to  VZ.     Draw  cv  per- 


FIG.  39 


pendicular  to  OC,  intersecting  ev  in  v.  Then  draw  DvO. 
Then  is  vO  equal  to  the  valve  travel.  If  the  middle  point, 
E,  of  vO  is  located,  and  >  circle  of  radius  EO  drawn  about  E 
as  a  center,  it  will  pass  through  c  and  e  as  well  as  O,  showing 
the  construction  to  be  correct. 


SLIDE-VALVES. 


PROBLEM  III. 

GIVEN.   Cut-off,  angle  of  lead,  width  of   port,  and  over- 
travel. 


FIG.  40 


REQUIRED.   Eccentricity,  lap,  lead,  and  angle  of  advance. 
SOLUTION.   Draw  XY  and  VZ,  Fig.  40,  at  right  angles  to 


VALVE-DIAGRAMS—GENERAL   PROBLEMS    OF  DESIGN.    $$ 

each  other.  Draw  OA,  the  crank  position  corresponding  to 
the  given  lead  angle;  that  is,  make  ZOA  equal  to  the  given 
angle.  Draw  OC,  the  crank-line  at  cut-off,  as  given.  Now, 
the  center  of  the  valve-circle  must  lie  somewhere  on  the  line 
bisecting  the  angle  formed  by  these  two  crank-lines.  There- 
fore draw  DOD' ',  bisecting  the  angle  CO  A,  and  the  angle 
XOD  is  the  angular  advance. 

Take  any  convenient  radius,  as  E 'O,  and  describe  a  trial 
valve-circle,  intersecting  OD  at  i/.  Then  Oc'  is  the  corre- 
sponding lap,  and  by  drawing  in  this  trial  lap-circle  it  will  be 
found  that  the  greatest  possible  opening  of  the  port  is  L'v1 ', 
with  this  lap  and  valve-travel.  The  greatest  possible  opening 
of  the  port  is  evidently  equal  to  the  width  of  the  port  plus 
the  overtravel,  and  as  both  of  these  are  given  in  the  problem, 
the  maximum  port-opening  is  known.  It  is  hardly  probable, 
though  possible,  that  the  distance  L'v'  will  equal  the  given 
amount  on  the  first  trial.  That  being  the  case,  proceed  by 
drawing  v'g  in  any  convenient  direction,  and  take  v'x  equal 
to  the  required  greatest  port-opening — that  is,  equal  to  the 
width  of  the  port  plus  the  overtravel.  Then  join  L'  and  x, 
and  from  O  draw  Og  parallel  to  L' x,  until  it  cuts  v'g  in  g. 
Then  is  v'g  equal  to  the  required  eccentricity.  Next  lay  off 
Ov  on  OD,  equal  to  v'g.  Bisect  Ov  in  E,  and  about  E  as  a 
center  and  with  a  radius  OE  draw  in  the  valve-circle  which 
will  cut  OC  at  Cj  thus  determining  the  lap,  Oc;  and  by  draw- 
ing the  lap-circle  the  lead  de  is  determined. 

The  maximum  port-opening  with  this  travel  and  lap  is 
evidently  vL\  and  if  the  construction  is  accurate,  vL  will  be 
found  to  be  equal  to  v'x. 

PROBLEM  IV. 

GIVEN.  Point  of  cut-off,  lead,  and  greatest  possible  port- 
opening. 

REQUIRED.   Lap,  valve-travel,  and  angle  of  advance. 


6o 


SLIDE-VALVES. 


SOLUTION.  This  is  shown  in  Fig.  41.  Draw  JTFand  VZ 
at  right  angles,  intersecting  at  O.  Draw  OCt  the  crank- line 
at  cut-off,  as  given.  Then  produce  OC  to  ;«,  making  Om 
equal  to  the  given  lead.  This  should  be  drawn  on  a  large 
scale,  as  the  lead  is  usually  small,  seldom  exceeding  T3^-  of  an 


inch.     A  scale  of  from  two  to  four  times  full  size  should  be 
employed. 

Next  take  ms  on  mC,  equal  to  the  given  maximum  port- 
opening.  Through  s  draw  nst  parallel  to  the  line  of  stroke 
VZj  making  st  equal  to  sm  as  shown  by  the  arc  mt.  Then 
join  O  and  t,  and  on  Ot  take  Oh  equal  to  Om  as  shown  by  the 


VALVE-DIAGRAMS—GENERAL   PROBLEMS   OF  DESIGN.   6 1 

arc  mh.  Next  about  O  as  a  center,  and  with  a  radius  equal  to 
On,  describe  an  indefinite  arc.  Through  h  draw  gh  parallel 
to  VZ.  Then  gh  will  cut  the  arc  just  drawn  in  g.  The  arc 
cuts  Ot  in  k,  and  this  arc  kn  is  next  picked  up  in  the  compass 
and  laid  off  from  g  to  D1 '.  Then  draw  D'O  and  produce  it 
indefinitely,  as  to  D.  Then  is  the  center  of  the  valve  circle 
on  D'OD,  and  DOX  is  the  angle  of  advance. 

The  problem  now  reduces  to: 

GIVEN.  Cut-off,  lead,  angle  of  advance,  and  greatest  port- 
opening. 

REQUIRED.  Eccentricity. 

This  is  the  same  as  Problem  III,  and  the  solution  is  made 
in  accordance  with  the  directions  given  for  that.  The  draw- 
ing, Fig.  41,  covers  this  solution,  and  is  lettered  the  same  as 
Fig.  40  to  enable  the  construction  to  be  followed  throughout, 
if  desired. 

The  proof  of  these  solutions  is  in  one  or  two  cases  rather 
mathematical,  and  is  therefore  omitted,  as  it  is  entirely  un- 
necessary to  understand  this  proof  in  using  the  diagrams. 

'In  order  to  fix  these  principles,  it  is  not  sufficient  to  rest 
content  with  merely  reading  the  foregoing  text,  but  it  is 
necessary  to  solve  numerous  examples  under  each  problem. 

Other  problems  will  probably  suggest  themselves  to  the 
mind,  but  they  can  either  be  reduced  to  the  preceding  ones, 
or  can  be  solved  by  the  application  of  the  general  principles 
of  the  diagram  as  outlined  in  Chapter  III. 


CHAPTER   VI. 
ROCKERS   AND    BELL-CRANKS. 

IT  has  been  said  that  the  motion  of  the  eccentric  is  very 
often  imparted  to  the  valve  through  a  rocker  which  joins  the 
end  of  the  eccentric- rod  to  the  end  of  the  valve-stem. 
This  is  a  necessary  appliance  when  the  path  of  the  valve 
is  not  in  the  same  straight  line,  that  is,  when  the  line  of 
valve-travel  is  parallel  to  the  line  of  stroke,  but  is  above  or 
below  it,  or  on  one  side  of  it. 

Figs.  42  and  43  show  two  forms  of  this  rocker,  the  prin- 
cipal difference  between  them  being  that  with  the  plain 
rocker,  Fig.  42,  where  the  pivot  is  at  the  bottom,  the  eccen- 


A  V    • 


FIG.  42 


trie-rod  at  the  top,  and  the  valve-stem  attached  somewhere 
between,  the  valve  motion  is  the  same  in  direction  as  that  of 
the  eccentric,  and  therefore  the  eccentric  precedes  the  crank, 

62 


ROCKERS  AND   BELL-CRANKS.  63 

as  shown  in  the  drawing;  while  with  the  bell-crank  shown  in 
Fig.  43,  with  the  pivot  at  the  angle,  the  motion  of  the  valve 
is  opposite  in  direction  to  that  of  the  eccentric,  arid  therefore 
the  eccentric  follows  the  crank.  In  both  cases  the  amount  of 
motion  of  the  valve  depends  upon  the  relative  lengths  of  the 


FIG.  43 


rocker-arms  to  which  the  valve-stem  and  eccentric-rods  are 
attached. 

The  method  of  laying  out  these  rockers  will  be  understood 
by  reference  to  the  figures,  which  are  lettered  alike,  as  the 
same  method  is  pursued  in  both  cases. 

Let  VZ  be  the  line  of  stroke,  and  O  the  center  of  the 
shaft.  Then  take  O  as  a  center,  and  with  a  radius  OE  equal 
to  the  eccentricity  describe  the  eccentric-circle,  as  shown. 
Let  the  direction  of  rotation  be  as  represented  by  the  arrow. 
Then  draw  inn  at  the  proper  distance  from  VZ  and  parallel  to 


64  SLIDE-  VA  L  VES. 

it,  to  represent  the  line  of  travel  of  the  valve-stem,  and  let  X 
be  the  position  of  the  end  of  the  valve-stem  when  the  valve 
is  in  its  middle  position.  Lay  off  XL  and  XL ',  each  equal  to 
the  outside  lap  of  the  valve.  Then  take  L  and  L'  as  centers, 
and  with  radii  equal  to  the  length  of  the  longer  arm  of  the 
rocker  describe  the  two  arcs  vs  and  xy,  intersecting  at  R. 
Then  is  R  the  pivot  of  the  rock-shaft.  Then  take  R  as  a 
center,  and  a  radius  equal  to  the  other  arm  of  the  rocker,  and 
draw  the  arc  st,  making  it  of  indefinite  length.  From  O  draw 
O  Y  tangent  to  the  arc  st,  determining  the  point  of  tangency, 
Yy  by  drawing  RY  perpendicular  to  OY,  the  point  of  intersec- 
tion being  the  required  point.  Next,  from  O,  draw  OE 
perpendicular  to  OY.  Then  OE  represents  the  eccentric,  and 
by  joining  E  and  Y  the  length  EY  is  determined,  which  is 
the  length  of  the  eccentric-rod. 

Peculiarities  of  the  design  may  make  it  an  object  to  make 
the  eccentric-rod  of  a  certain  length.  This  may  be  done  by 
shifting  the  pivot  of  the  rocker  until  the  desired  length  is 
obtained.  If  the  distance  through  which  the  pivot  is  to  be  so 
moved  is  short,  proceed  as  follows: 

Through  Y  draw  a  line  YG  parallel  to  the  line  of  stroke 
VZ.  With  E  as  a  center,  and  with  a  radius  Ea  equal  to  the 
given  length  of  the  eccentric  rod,  describe  an  arc  ah,  intersect- 
ing YG  in//.  Then  move  the  whole  arrangement  through 
the  distance  YH,  changing  the  length  of  the  valve-stem  by 
that  amount.  The  same  construction  would  hold  good  if  the 
rod  were  to  be  shortened  instead  of  lengthened. 

With  this  construction  the  action  of  the  valve  is  not  made 
irregular  at  either  admission  or  cut-off.  The  angle  between 
the  crank  and  eccentric  is  changed,  but  this  is  a  matter  which 
affects  the  valve-setting  alone.  The  amount  of  the  change 
may  be  found  by  laying  off  ZOC  equal  to  the  angle  of  advance 
as  found  by  the  valve  diagram,  and  then  OC  represents  the 
position  of  the  crank  with  the  valve  in  its  middle  position,  and 
COE  is  the  angle  between  the  crank  and  eccentric. 


ROCKERS  AND    BELL-CRANKS.  6$ 

Sometimes  the  valve-face  is  not  parallel  to  the  line  of 
stroke.  This  is  a  combination  seldom  met,  as  the  lack  of 
parallelism  renders  it  a  difficult  matter  to  face  off  the  valve- 
seat  and  bore  the  cylinder.  Therefore  an  extended  descrip- 
tion of  this  type  will  be  omitted.  There  is  one  special  case, 
however,  which  it  is  well  to  bear  in  mind,  and  that  is  when  the 
center  line  of  the  valve-stem  is  inclined  to  the  line  of  stroke 
in  such  a  manner  that  it  would,  if  produced,  pass  through  the 
center  of  the  shaft.  This  changes  the  angle  of  advance  by 
just  the  angle  between  the  two  lines — the  line  of  stroke  and 
the  center  line  of  the  valve-stem.  If  the  change  is  but  slight, 
it  may  be  neglected.  In  any  case,  it  is  a  matter  which  affects 
the  valve-setting  alone. 

In  both  Figs.  42  and  43  the  rocker-arms  are  unequal,  the 
shorter  arm  being  f  of  the  larger  one,  the  resultant  travel  of 
the  valve  being  the  same  as  if  the  eccentricity  were  £  of  OE+ 
In  designing  and  laying  out  the  valve,  it  is  treated  as  if  it 
were  actuated  by  the  larger  eccentric.  That  is  to  say,  no 
allowance  is  made  in  construction  until  it  comes  time  to  lay 
out  the  eccentric,  when  that  is  made  with,  in  this  case,  f  of 
the  throw  which  would  be  given  it  if  it  were  either  connected 
direct  or  with  an  equal-armed  rocker. 

The  valve  diagrams  given  in  Chapters  III  and  V  show  that 
if  the  cut-off  is  equalized  on  the  two  ends — that  is,  if  made 
to  occur  at  the  same  point  on  each  stroke — by  making  the  out- 
side laps  unequal,  the  leads  will  be  unequal;  and  if  the  leads 
are  kept  equal,  the  cut-offs  will  be  unequal.  It  is  possible, 
however,  by  the  use  of  a  bell-crank  designed  in  a  certain  way, 
to  equalize  the  cut-offs  and  have  the  leads  equal  or  nearly  so. 
This  bell-crank  is  designed  as  follows:  Determine  the  outside 
lap,  angle  of  advance,  and  eccentricity  necessary  to  secure 
the  required  cut-off  and  lead  on  one  end  of  the  valve.  It 
makes  no  difference  which  end  of  the  cylinder  is  selected,  as 
the  difference  between  the  laps  would  be  compensated  for  by 
the  difference  between  the  resulting  bell-cranks.  Then  pro- 


66 


SLIDE-VALVES. 


ceed  to  construct  the  diagram  shown  in  Fig.  44,  which  should 
be  drawn  on  as  large  a  scale  as  possible  in  order  to  render  the 
work  accurate. 

Draw  VZ  to  represent  the  line  of  stroke,  and  let  O  be  the 
center   of  the  crank-shaft.     About  O  as  a  center,  and  with 


/,'       T    VALVE  STEM 


a  radius  equal  to  the  length  of  the  crank,  describe  the 
crank-circle.  Then  describe  the  eccentric-circle  about  O  as  a 
center  and  with  a  radius  Oc  equal  to  the  eccentricity.  Next 
find  the  crank  positions  corresponding  to  the  given  point  of 
cut-off  on  the  forward  and  return  strokes  respectively.  These 
are  OC  and  OC' ,  and  may  be  found  by  either  of  the  two 
methods  given  in  Chapter  II.  The  first  method  is  the  one 
adopted  here,  because  the  method  of  Deprez,  requiring  the 
use  of  the  auxiliary  crank-circle,  would  complicate  the  draw- 
iftg  too  much  at  the  right-hand  end.  Next  draw  in  OA  and 
OA',  the  crank-positions  at  admission  on  the  forward  and 
return  strokes  respectively.  These  positions  are  determined 
by  taking  the  crank-position  for  admission  from  the  end  of 
the  cylinder  for  which  the  lap,  etc.,  were  determined,  and 
making  the  other  lead  angle  the  same.  For  instance,  if  OA 
was  found,  then  the  angle  ZOA'  is  made  equal  to  VOA.  The 
next  step  is  to  locate  the  eccentric-positions  corresponding  to 
these  four  crank-positions.  This  is  done  by  laying  off  from 
each  crank-position  the  angle  between  the  crank  and  eccentric. 
That  is,  COc,  C'Oc'AOa,  and  A' Oa'  are  each  equal  to  the 
angle  between  the  crank  and  eccentric. 

Now,  it  must  be  remembered  that  when  admission  takes 


ROCKERS  AND    BELL-CRANKS.  6? 

place,  the  valve  is  just  opening  the  port;  and  when  cut-off 
occurs,  the  valve  is  just  closing  the  port.  Consequently  the 
valve  is  in  the  same  position  at  admission  and  cut-off,  but  the 
motions  are  in  opposite  directions.  That  is  to  say,  with  the 
eccentric-rod  end  at  either  a  or  c  the  valve  must  be  in  the 
same  position.  Then  with  a  and  c  as  centers,  and  with  radii 
equal  to  the  length  of  the  eccentric-rod,  describe  arcs  inter- 
secting at  5.  Then  take  the  same  radius,  and  with  a'  and 
c'  as  centers  describe  two  more  arcs  which  intersect  at  t. 
Join  s  and  /,  and  find  the  middle  point  x.  At  x  draw  xR, 
perpendicular  to  ts.  Then  if  a  line  be  drawn  perpendicular 
to  the  line  of  stroke,  as,  for  example,  RT,  it  will  determine 
the  angle  xR  T  between  the  two  arms  of  the  bell-crank.  The 
lengths  of  these  two  arms  must  be  such  as  to  give  the  required 
movement  of  the  valve,  and  are  determined  as  follows. 

On  xR  lay  off  xm  equal  to  xs,  and  from  m  draw  the  line 
mn  perpendicular  to  the  line  of  travel  of  the  valve-stem,  and 
make  mn  equal  to  the  lap  of  the  valve.  Draw  xT  from  x 
through  n  until  it  meets  the  line  of  travel  of  the  valve-stem 
at  T;  and  from  T  draw  TR,  also  perpendicular  to  the  line  of 
valve-stem  travel  until  it  meets  xR  at  R.  Then  will  TR  and 
Rx  be  the  lengths  required  for  the  two  arms  of  the  bell- 
cranks,  while  R  will  be  the  point  of  suspension.  The  valve- 
stem  is  fastened  to  T,  and  the  eccentric-rod  to  x.  The 
general  appearance  is  shown  by  the  heavy  lines. 

In  a  similar  manner  the  exhaust  and  compression  can  be 
equalized,  obtaining  points  similar  to  /  and  s\  and  if  these 
points  should  happen  to  lie  in  such  positions  that  an  arc  pass- 
ing through  them  would  also  pass  through  s  and  /,  it  would 
be  possible  to  design  a  bell-crank  which  would  equalize  all  the 
events  of  the  stroke.  As  an  ordinary  thing  this  coincidence 
is  impossible,  so  it  may  be  taken  as  an  accepted  fact  that  with 
a  plain  D  valve  it  is  impossible  to  equalize  all  the  events  of 
the  stroke.  This  is  one  of  the  greatest  drawbacks  to  this  type 
of  valve. 


CHAPTER  VII. 
DESIGN   OF   A   PLAIN   D-VALVE. 

IF  the  preceding  chapters  have  been  thoroughly  mastered, 
the  student  is  in  possession  of  sufficient  information  to  design 
a  plain  slide-valve.  In  order  to  fix  clearly  the  various  steps 
and  their  order,  a  numerical  example  will  be  worked  out 
completely. 

Let  the  engine  be  24  X  24,  running  at  135  revolutions  per 
minute.  Cut-off  to  be  at  $  stroke,  compression  at  T*L  stroke, 
and  the  lead  -J  of  an  inch.  The  connecting-rod  is  five  times 
the  length  of  the  crank.  Length  of  port  22  inches. 

ORDER   IN  WHICH   THE  VARIOUS    DIMENSIONS  ARE   TO   BE 
DETERMINED. 

T  3  £    "V    2  A.    y    2 

1.  Piston  Speed. — This  is  -       -  =  540  feet    per 

minute. 

2.  Area  of  Steam-port. — This  is  determined  by  reference 
to  Table  III,  assuming  that  the  velocity  of  steam  at  exhaust 
is  4000  feet  per  minute.      Under  *  *  4000,  Area  of  Port ' '  and 
opposite  550,  which  is  the  nearest  to  the  given  piston  speed, 
is  .138.     The  piston  being  24  inches  in  diameter,  its  area  is 
452.39  square  inches,  and  the  port  area  is  therefore 

452.39  X  .138  =  62.430  square  inches. 

3.  Width   of  Steam-port. — This  is   the    area   just   found 
divided  by  the  length  given,  or 

62.430  -r-  22  =  2.837  inches. 

68 


DESIGN  OF  A    PLAIN  D-VALVE.  69 

The  nearest  sixteenth  is  2-J-J,  from  Table  II,  which  is  there- 
fore the  width  required. 

4.  Maximum    Port-opening.  —  This  will    be   the    amount 
which  would  allow  the  entering  steam  to  have  a  velocity  of 
6000  feet  per  minute.      It  may  be  found  by  taking  the  num- 
ber from  Table  IV,  multiplying  it  by  the  piston  area,   and 
then  dividing  it  by  the  port  length;  but  it  is  easier  to  multiply 
the  result  given  in  (3)  by  f ,  as  4000  is  f  of  6000.     This  gives 

2.837  X  f—  1.891  inches, 

and,  from  Table  II,  the  maximum  port-opening  is  made  i-J 
inches. 

5.  Outside  Lap,  Angle  of  Advance,    Valve-Travel. — These 
must  be  such  as  to  produce  the  required  point  of  cut-off  with 
the  given  amount  of  lead,  and  the  port-opening  found  in  (4). 
This    amounts    to    an    example    in    Problem     IV,    given    in 
Chapter  V. 

First  determine  the  crank-position  corresponding  to  the 
given  point  of  cut-off  on  either  end — the  head-end,  for 
example.  This  is  done,  as  shown  in  Fig.  45,  by  the  method 
given  in  Chapter  II.  For  this  purpose  a  sheet  of  paper  is 
required  which  will  allow  the  drawing  to  be  made  on  a  large 
enough  scale  to  measure  accurately.  The  crank  circle  being 
24  inches  in  diameter,  a  sheet  of  paper  fourteen  inches  square 
will  be  large  enough  for  a  half-size  drawing.  Then  draw  the 
two  lines  XY  and  VZ  at  right  angles,  locating  O,  which  will 
represent  the  center  of  the  shaft.  Then  locate  O',  the 
center  of  the  auxiliary  crank-circle.  The  connecting-rod 
being  five  times  as  long  as  the  crank,  OO'  will  be  ^  of  the 
crank,  as  shown  in  Table  I.  That  is,  OO'  will  be 

—  =  -  =  .6  of  an  inch, 
20       5 

and  on  the  half-size  drawing  it  will  be  .3  of  an  inch.     Then 


SLIDE-VALVES. 


draw  in  the  two  circles  shown,  the  dotted  circle  being  the 
auxiliary  crank-circle.  Lay  off  vn  equal  to  -J-  of  vz,  as  the 
cut-off  is  to  be  at  -J  stroke.  Draw  nc  perpendicular  to  vz, 
cutting  the  auxiliary  circle  at  c,  and  join  O  and  c,  producing 


it  until  it  meets  the  crank  circle  at  C.  Then  OC  is  the  crank 
position  at  cut-off  on  the  head  end.  (Refer  to  Chapter  II 
for  a  complete  explanation  of  this  method.) 

Now,  having  located  the  crank-position  at  cut-off,  take 
another  sheet  of  paper,  and  proceed  as  in  Problem  IV. 
Draw  XY  and  VZ  at  right  angles,  Fig.  46.  Their  intersec- 
tion represents  the  center  of  the  shaft.  Draw  OC,  the  crank- 


DESIGN  OF  A    PLAIN  D -VALVE.  ? I 

position  at  cut-off,  as  found  in  Fig.  i.  Produce  OC  to  m, 
making  Om  equal  to  the  given  lead,  j-  of  an  inch.  This 
drawing  should  be  made  at  least  twice  full  size,  in  which  case 
a  sheet  of  paper  12  inches  square  will  be  large  enough.  Next, 
on  mC  take  ms  equal  to  I-J  inches,  the  given  maximum  port- 


opening.  Through  s  draw  nst  parallel  to  the  line  of  stroke 
VZ,  making  st  equal  to  sm  or  I-J  inches,  as  shown  by  the  arc 
mt.  Then  join  O  and  /,  and  on  Ot  take  Oh  equal  to  Om  or 
•J-  inch,  as  shown  by  the  arc  mk.  Then  about  O  as  a  center, 
and  with  a  radius  equal  to  On,  describe  an  indefinite  arc. 
Through  h  draw  gh  parallel  to  VZ.  Then  gh  will  cut  the 
indefinite  arc  just  drawn  in  g.  The  arc  cuts  Ot  in  k.  Then 
pick  up  the  arc  kn  in  the  compass  or  dividers,  and  lay  it  off 
from  g  to  D'\  that  is,  kn  equals  gD1 .  Then  join  D'  and  O, 
and  produce  D' O  to  any  point,  as  D.  Then  is  DOD'  the 
center  line  of  the  valve  circles,  and  DOX  is  the  angle  of 


72  SLIDE-VALVES. 

advance.  Now  take  any  convenient  radius,  as  E'O,  and 
describe  the  trial  valve-circle,  intersecting  OD  at  v' .  Then 
Oc'  is  the  corresponding  lap,  and  by  drawing  in  this  trial  lap- 
circle  it  will  be  found  that  the  greatest  possible  port-opening 
is  L'v'  with  this  lap  and  valve-travel.  If  this  is  not  equal  to 
i-J  inches,  the  given  amount — and  it  probably  will  not  be, — 
draw  v'y  in  any  convenient  direction,  and  take  v'x  equal  to  i-J 
inches,  the  required  maximum  port-opening.  Then  join  L' 
and  x,  and  from  O  draw  Oy  parallel  to  L '  x,  until  it  cuts  v'y 
in  y.  Then  is  v'y  the  required  eccentricity.  Lay  off  Ov 
equal  to  v'y,  and  draw  in  the  valve-circle  with  this  diameter, 
which  will  be  found  to  be  2-J  inches.  This  valve-circle  will 
cut  OC,  the  crank-position  at  cut-off,  at  c,  and  Oc,  which  is 
ij1^  inches,  is  the  outside  lap  on  the  head-end.  The  lead  is 
de,  and  it  will  be  found  to  measure  \  of  an  inch. 

6.  Inside  Lap,  Head- End. — This  could  be  found  on    the 
same  sheet,  but  by  this  time  that  drawing  has  become  too 
complicated  and  full  of  lines  to  permit  of  any  close  determina- 
tions.     Take  another  sheet  of  paper,  and  proceed  to  draw  the 
full  diagram  for  the  head-end,  as  shown  in  Fig.   47.     This 
drawing  may  be  made  on  a  smaller  scale  than  the  preceding 
one,  if  desired,  but  it  is  not  desirable  to  make  it  any  less  than 
actual  size.      Lay  off  the  things  already  determined  on  this 
sheet— that  is,   draw  XY,   VZ,  DOD' ,  OC,  the  upper  valve- 
circle  whose  center  is  E,  and  the  outside  lap-circle.       Now 
turn  back  to  Fig.  45,  and  locate  thereon  the  crank-position 
O T,  the  crank-position  corresponding  to  the  given  point  of 
compression  on  the  head  end.     (Compression  occurs  on  the 
return  stroke,  and   the  crank  is  therefore  below  the  line  of 
stroke  when  it  occurs.)     Transfer  OT  to  Fig.  47,  where  it  will 
intersect  the  lower  lap  circle  at  /,  thus  determining  Ot,  the 
inside  lap  on  the  head  end,  which  will  be  found  to  measure  -J 
an  inch.     Release  and  admission  will  take  place,  as  shown,  at 
the  crank-positions  OA  and  OR. 

7.  Inside  and  Oiitside  Laps  on  Crank-End. — These  can  be 


DESIGN  OF  A    PLAIN-  D-VALVE. 


73 


V 


74  SLIDE-  VA  L  VES. 

found  very  readily  on  the  last  drawing,  but  for  purposes  of 
explanation  they  are  determined  separately  in  Fig.  48,  which 
is  an  exact  reproduction  of  Fig.  47  as  far  as  the  angle  of 
advance  and  the  valve-circles  are  concerned.  Then  pass  back 
to  Fig.  45  and  determine  OC'  and  OT',  the  crank-positions 
corresponding  to  cut-off  and  compression  on  the  crank-end. 
Then  transfer  them  to  Fig.  48,  and  thus  determine  Oc' ,  the 
outside  lap,  to  be  -J^  inch,  and  Of ,  the  inside  lap,  which  is  |- 
of  an  inch. 

Of  course  in  making  these  determinations,  all  the  crank- 
positions  required  will  be  first  located  on  Fig.  45  and  marked 
for  identification,  and  then  transferred  to  Figs.  46  and  47  as 
needed.  Fig.  48  will  be  dispensed  with. 

8.  Minimum  Width  of  Bridge. — In  this  case  the  valve  has 
no  overtravel,  and  the  bridge  width  should  be  made  equal  to 
the  thickness  of  the  cylinder-walls,  which  are  assumed  in  this 
case  to  be  one  inch. 

9.  Width  of  Exhaust-port. — This  is  determined   by   the 
method   given   in    the   preceding  chapter.      There   are   three 
different  rules  given  there,  and  in  addition  to  that  there  is  a 
difference  between  the  two  ends  of  the  valve,  thus  giving  rise 
to  the  question  as  to  which  is  the  proper  one   to  be  used. 
Always  employ  the  one  which  gives  the  greatest  result,  thus 
being  sure  that  the  port  is  plenty  wide  enough.      In  this  case 
the  third  rule  is  the  one  to  be  taken,  giving 

Half-travel  +  port  width  —  bridge  -f  inside   lap  =  exhaust-port    width. 

2*          +         2T\  '       +  t  5TV 

The  results  obtained  on  the  preceding  text  may  be  col- 
lected as  follows: 

Eccentricity   2  J- 

Valve  travel  (—  2  X  eccentricity) 5 j- 

Head  end,  outside  lap   i-^ 

"  inside  lap J- 


DESIGN  OF  A    PLAIN  D-VALVE. 


FIG.  48 


SLIDE-VALVES. 


Crank-end,  outside  lap 

'  '  inside  lap 

Exhaust-port  width 

The  next  thing  in  order  is 


TO    LAY    OUT   THE   VALVE. 

This  is  a  comparatively  simple  matter,  but  one  in  which 
many  beginners  go  astray,  owing  to  the  fact  that  the  valve  is 
not  the  same  on  both  ends,  and  consequently  cannot  be  laid 
out  very  well  from  a  center  line. 

Draw  the  line  FZ,  Fig.  49,   to  represent  the  valve-seat. 


Then  start  to  lay  out  the  valve  from  either  end — the  crank 
end  being  taken  in  this  case.  Start  at  any  point,  as  A,  and 
lay  out  the  valve  as  follows: 

Make  AB  =  outside  lap,  crank-end  —  IT3F  inches. 
BC  =  width  of  port  =  2^\       " 

CE  =       "      ll  bridge  "=  I  " 

EP—       «      «  exhaust 
FH=       ll      "  bridge  =  I 

HI—       li     "  port  = 

//=  outside  lap,  head-end    = 


DESIGN  OF  A    PLAIN  D-VALVE.  77 

Then  draw  in  the  section  of  the  ports  as  shown,  making 
the  thickness  of  the  seat  one  inch,  the  same  as  the  cylinder- 
walls  and  bridge.  Next, 

Make  CD  =  inside  lap,  crank  end  =  £•  inch. 
HG  =  inside  lap,  head  end    —  ^     " 

Then  draw  in  the  valve  proper,  making  it  thick  enough  to 
withstand  the  pressure  to  which  it  is  subjected.  The  height 
of  the  cavity  in  the  valve  need  not  be  very  great,  the  general 
proportions  shown  in  the  cut  being  very  good. 

Laying  out  the  valve-seat  first  and  the  valve  afterwards 
avoids  confusion. 


CHAPTER   VIII. 

LENGTH   OF  VALVE-CHEST,   VALVE-STEM,  AND    ECCEN- 
TRIC-ROD. 

IN  addition  to  the  dimensions  of  the  valve  and  seat  as 
determined  in  the  preceding  chapter,  there  are  three  more  to 
be  found.  These  are: 

1.  The  length  of  the  valve-chest. 

2.  The  length  of  the  valve-stem. 

3.  The  length  of  the  eccentric-rod. 

These  will  be  considered  in  the  order  given. 


LENGTH    OF   VALVE-CHEST. 

This  must  be  at  least  long  enough  to  allow  the  full 
travel  of  the  valve.  That  is,  it  must  permit  the  valve  to  go 
to  the  two  extremes  of  its  travel  as  shown  in  Figs.  50  and  51. 


The   minimum   length  of  the  chest  may  then    be  measured 
directly  from  the  drawing,  which  represents  the  valve  in  its 

78 


LENGTH  OF   VALVE-CHEST,    VALVE-STEM,  ETC.          Jg 

middle  position,  as  in  Fig.  49,  by  laying  off  the  half-travel  at 
each  end  of  the  valve,  and  measuring  between  the  marks 
thus  obtained.  Or  if  it  is  desired  to  obtain  the  length  of 


valve-chest  necessary  without  drawing  the  figure,  it  can  be 
done  by  applying  the  following  rule: 

Add  together  the  outside  laps  at  the  two  ends  of  the  valve > 
twice  t/ie  width  of  the  steam-port,  twice  the  thickness  of  the 
bridge,  the  'cvidth  of  the  exhaust-port  and  the  valve-travel. 
This  gives  the  minimum  inside  length  of  the  steam-chest ',  and  it 
should  be  increased  by  at  least  one  inch. 


LENGTH   OF   VALVE-STEM. 

One  of  the  most  common  methods  of  connecting  the  valve 
and  valve-rod  or  stem  is  shown  in  Fig.  52.  This  consists  of 
a  collar  which  fits  closely  around  the  valve,  and  may  or  may 
not  be  otherwise  secured  to  it.  The  collar  is  tapped  at  B,  and 
the  valve-stem  screwed  in.  The  length  of  the  valve-rod,  VR, 
is  measured  from  the  inside  of  the  collar  to  the  point  at 
which  the  stem  is  secured  to  the  eccentric-rod,  or,  if  a  rocker 
is  employed,  to  that.  The  length  of  the  valve-stem  must  be 
sufficient  to  permit  its  reaching  from  the  valve  when  in  its 
extreme  position,  as  shown  in  Fig.  51,  to  the  point  where 


8c 


SLIDE-VALVES. 


it  joins  either  the  eccentric-rod  or  the  rocker-arm.  This 
gives  a  direct  means  of  obtaining  the  proper  length  from  the 
drawing.  The  valve  shown  in  the  drawing  has  overtravel. 


LENGTH    OF   THE    ECCENTRIC-ROD. 

Figs.    42,    43,   and    44    in    the    sixth    chapter  show    the 
determination  of  this  length  when  a  rocker  or  bell-crank  is 


Fia.  52 

employed.     When  neither  is  used,  the  length  of  the  eccentric- 
rod  may  be  found  by  this  rule: 

Find  the  distance  from  the  center  of  the  exhaust-port  to  the 
center  of  the  shaft.  This  may  be  done  from  the  drawing. 
From  that  distance  subtract  the  distance  from  the  center  of  the 
exhaust-port  to  the  end  of  the  valve -stem,  when  the  valve  is  in 
its  middle  position.  The  remainder  is  the  required  length  of 
the  eccentric-rod. 

This  is  not  strictly  true,  but  it  is  near  enough,  the  differ- 
ence between  the  length  thus  obtained  and  the  correct  length 
being  negligible,  as  will  be  apparent  from  Fig.  53.  O  is  the 


LENGTH  OF   VALVE-CHEST,    VALV. 


81 


center  of  the  shaft,  P  is  the  center  of  the  exhaust-port,  and 
PV  is  the  length  of  the  valve-stem.      The  length  given  by  the 


FIG.  53 


rule  is  OV,  whereas  the  correct  length  would  be  EP.  But, 
for  example,  suppose  OE  were  3  inches  and  O  V  were  5  feet 
or  60  inches.  Then 


=  1/3600  -|-  9 

=  1/3609 

=  60.07  inches. 

The  correct  length  is  60.07  inches,  then,  against  60 
obtained  by  the  rule. 

The  rules  for  the  lengths  of  valve-stem  and  eccentric-roxl 
are  given  in  such  a  way  as  to  permit  either  one  to  be  assumed 
and  the  other  one  found.  The  location  of  the  point  of  sus- 
pension of  a  rocker,  if  one  is  employed,  usually  determines 
both  of  these  lengths. 


CHAPTER    IX. 
DESIGN   OF  AN  ALLEN  OR  "TRICK"  VALVE. 

THE  difficulty  with  the  ordinary  slide-valve  is  that  it  can- 
not be  used  for  early  cut-offs,  on  account  of  the  large  outside 
laps  and  travel  required  to  produce  the  desired  result.  This 
renders  the  friction  of  the  valve  a  very  considerable  item, 
and  it  is  unusual  to  find  a  plain  slide-valve  used  to  secure  cut- 
offs much  earlier  than  five-eighths  of  the  stroke,  three-quarters 
being  usually  the  limit  for  which  they  are  employed. 

There  are,  however,  several  valves  constructed  on  the 
same  general  principle,  which  permit  an  early  cut-off  to  be 
attained  with  moderate  lap  and  travel.  Prominent  among 
these  is  the  Allen,  or  "  Trick,"  valve,  shown  in  Figs.  54,  55, 
and  56,  which  is  so  constructed  as  to  give  double  the  amount 
of  port-opening  as  a  plain  slide  with  the  same  amount  of 
travel,  or  the  same  port-opening  with  one  half  the  travel. 
This  advantage  is  secured  by  means  of  the  passage  X  in  the 
valve.  This  passage  is  used  for  admission  only. 

While  this  valve  is  designed  to  take  the  place  of  a  slide- 
valve,  it  cannot  be  used  on  the  same  seat,  but  requires  the 
face-plate  shown  between  the  slide-valve  seat  and  the  face  of 
the  Allen  valve.  Its  operation  will  be  best  understood  by 
reference  to  the  figures. 

The  figures  show  the  valve  in  its  progress  from  one 
extreme  of  its  travel  to  the  other,  and  the  arrows  show  the 
course  of  the  steam.  Fig.  54  represents  the  extreme  left- 
hand  position,  Fig.  55  shows  the  middle  position,  and  in  Fig. 
56  is  shown  the  extreme  right-hand  position.  With  the 

82 


DESIGN  OF  AN  ALLEN  OR   "  TRICK"    VALVE.  83 

valve  out  at  the  left  the  piston  is  over  toward  the  right,  and 
steam  must  be  admitted  to  the  right-hand  end  of  the  cylinder 
through  the  port  P.  This  is  open  by  the  amount  BE, 
Fig.  54,  and  the  steam  enters  through  that  space  just  as  it 


would  with  an  ordinary  D-valve.  But,  in  addition,  the  other 
end  of  the  valve  has  moved  beyond  the  edge  of  the  face-plate, 
putting  the  passage  X  in  communication  with  the  steam-chest, 
through  the  opening  C 'D' r,  and  steam  is  therefore  admitted  to 
the  right-hand  end  of  the  cylinder  through  CD.  The  total 
opening  of  the  port  is  therefore  the  combined  amount  of  CD 
and  BE.  If  CD  and  BE  are  equal,  the  total  port-opening  is 


84  SLIDE-  VA  L  VES. 

2BE-,  and  as  BE  is  the  amount  that  the  port  would  be  open 
with  a  plain  slide-valve,  it  is  evident  that  the  Allen  valve 
secures  double  the  port-opening  with  the  same  travel.  The 
action  at  the  other  end  of  the  valve  is  the  same. 

-Another  point  which  must  be  borne  in  mind  in  regard  to 
this  valve  is  that  it  cannot  be  used  to  advantage  when  the 
cut-off  is  much  later  than  one-half  stroke. 

In  order  to  bring  out  clearly  one  or  two  points,  an  actual 
valve  will  be  designed. 

PROBLEM.  To  lay  out  an  Allen  valve  and  seat  for  a 
12  X  24  engine,  running  at  150  revolutions  per  minute,  con- 
necting-rod four  times  the  crank,  cut-off  on  both  ends  at  10 
inches,  compression  at  22  inches,  lead  on  head-end  -J  inch. 
Ports  to  be  1 1  inches  long. 


ORDER   IN   WHICH    THE   DIMENSIONS   ARE   DETERMINED, 


I.  Piston  Speed. — This  is,  by  the  rule  given  before, 


24 
2  X  150  X  —  =  600  feet  per  minute. 


2.  Area  of  Port. — From  Table  IV  this  is  found  to  be 
.15   of  the  area  of  the  piston,   if  the  velocity  of  steam  at 
exhaust  is  4000  feet  per  minute.      The  area  of  the   12-inch 
piston  is  113.10  inches,  so  that  the  port  area  must  be 

113.10  x  .15  =  !6.95  inches. 

3.  Width  of  Port. — This  is  the  area  divided  by  the  length* 

16.95  -*-  ii  —  1.54  inches, 
or  from  Table  II,  ITV  inches. 


DESIGN  OF  AN  ALLEN  OR   "  TRICK"    VALVE.  8$ 

4.    Maximum  Port-Opening.  —  The   velocity  of    steam    at 
admission  being  assumed  at  6000,  the  port-opening  required  is 

4000 

X  iA  =  % 


and  therefore,  according  to  Table  II,  it  is  made  IT\  inches. 

5  .  Outside  Lap,  A  ngle  of  Advance,  and  Valve-  Travel.  —  This 
means  the  outside  lap,  angle  of  advance,  and  valve-travel 
necessary  to  give  the  required  point  of  cut-off  on  the  head- 
end for  an  ordinary  D-  valve,  but  with  one-half  the  given  port- 
opening.  The  Allen  valve  will  double  this  half,  giving  the 
required  amount. 

First  determine  the  crank-position  corresponding  to  cut- 
off on  the  head-end.  This  operation  has  been  fully  explained 
in  previous  chapters,  and  will  not  be  touched  on  again. 

Having   located   this   crank-position,   the    example   then 
becomes  a  case  of  Problem  IV,  Chapter  V.     The  diagram  is 
given  in  Fig.  57,  and  the  dimensions  determined  therefrom 
are: 

Valve  travel  ...................     4^  inches. 

Eccentricity  (=  half-travel)  ......  2j 

Outside  lap,  head-end  ........  ....   if       " 

6.  Inside  Lap,  Head-End.  —  This  is  determined  from  the 
diagram,  Fig.  58,  where  T  is  the  crank-position  with  the 
piston  at  22  inches.  The  inside  lap  is,  as  explained  in 
Chapter  III,  that  part  of  the  crank-position  included  between 
the  valve-circle  and  the  point  O  of  the  diagram.  The  inside 
lap  and  outside  lap  are  read  from  different  valve-circles.  It 
will  be  noticed  that  the  line  O  T  does  not  cut  the  lower  valve- 
circle  below  the  line  of  stroke,  but  above  it.  The  inside  lap 
is  therefore  negative  ;  that  is,  the  valve  has  inside  clearance. 
While  this  is  a  possible  construction,  it  is  not  at  all  desirable, 
as  there  would  evidently  be  a  time  in  the  stroke  when  the  two 
ends  of  the  cylinder  would  be  in  direct  communication  with 


86 


SLIDE-  VAL  VES. 


each  other  and  with  the  exhaust.  This  will  be  plain  on  refer- 
ence  to  Fig.  55,  where  the  dotted  lines  x x'  show  the  construc- 
tion of  a  valve  with  inside  clearance  at  both  ends.  If  only  one 
end  has  this  negative  inside  lap,  the  blow-through  of  steam 
would  not  be  in  evidence  with  the  valve  in  its  middle  posi- 
tion, but  will  appear  later,  unless  the  other  end  has  a  con- 


FIG.  57 


siderable  amount  of  inside  lap,  at  least  equal  to  the  inside 
clearance  of  the  first  end. 

The  inside  lap  on  the  head-end  will  therefore  be  made  zero. 

7.  Inside  and  Outside  Laps,  Crank- End. — These  are  found 
from  the  diagram,  in  the  manner  already  explained.  The 
outside  lap  is  Oc' ',  Fig.  58,  which  is  IT5^-  inches.  The  inside 
lap  is  again  negative,  as  shown  by  OT'  cutting  the  upper 
valve-circle  below  the  line  of  stroke.  The  inside  lap  will 
therefore  be  made  zero. 


DESIGN  OF  AN  ALLEN   OR   "  TRICK"    VALVE.  «7 

For  the  rest  of  the  dimensions  the  description  will  refer  to 
the  valve  in  its  middle  position,  as  in  Fig.  55. 

8.  The  Distances  DE  and  D' E' . — These  are  equal  and  may 
be  assumed,  and  in  this  case  will  be  made  -J  inch. 

9.  Width  of  Port,  MN  and  M' N' . — These  being  the  open- 


ings  in  the  cylinder-face,   they  are  made  equal  to  the  port 
width  determined,  or  IT9F  inches  each. 

10.  Width  of  Passage  in  Valve,  or  CD  and  C'D'. — These 
are  each  made  equal  to  one-half  the  port-opening.     The  port- 
opening  being  i^,  the  half  is  ii|-,  and  CD  and  C'D'  will  be 
made  the  next  larger  sixteenth,  or  T9¥. 

These  last  three  are  all  the  dimensions  that  are  the  same 
on  both  ends  of  the  valve. 

1 1.  BE  is  made  equal  to  the  outside  lap  on  the  head-end, 
assuming  the  cylinder  to  be  at  the  right.     BE  =  if  inches. 


88  SLIDE-  VA  L  VES. 

12.  B ' E'  is  made  equal  to  the  outside  lap  on  the  crank- 
end,  or  IT5F  inches. 

13.  AC.     This  is  equal  to  the  outside  lap,  head-end,  ij 
inches. 

14.  A'C'.     This  is  equal  to  the  outside  lap  on  the  crank- 
end,  iT5g-  inches. 

15.  BC.     This  is  made  equal  to  the  outside  lap,  head-end 
— one-half  the  port-opening — DE.     This  gives  if  —  T9^  —  -J 
=  i-jig-  inches. 

16.  B'C'.     This  is  made  equal  to  the  outside  lap,  crank 
end — one-half  the  port-opening — D'E'.     This  gives  IT5¥  —  T9^ 
—  -J  —  -|  inches. 

No  allowance  is  made  for  the  difference  between  the  port- 
openings  at  the  two  ends  of  the  cylinder,  because  the  crank- 
end  opening,  as  shown  by  the  diagram,  is  greater  than  that 
of  the  head-end.  Therefore,  if  the  dimensions  be  made  large 
enough  to  permit  the  required  opening  on  the  head-end,  the 
crank-end  opening  will  be  ample  with  the  same  allowance. 

The  next  thing  is  to  describe  the  method  of 

LAYING   OUT   THE   FALSE   SEAT   AA'. 

17.  Thickness. — This  may  be  made  anything  desirable. 

1 8.  AE.     This  is  made  equal  to  the  outside  lap,  head-end 
-\-  one  half  the  port-opening  -f-  DE\  or,  in  this  case,  if  -f-  -£$ 
+  i=  2^  inches. 

19.  A'E'.     This  is  calculated  in  a  similar  manner:  Out- 
side lap,  crank-end  +  one  half  of  the  port-opening  +  D'E', 
which  gives  IT^  +  yV  4~  t  ~  2  mcries. 

20.  EH.     This  made  equal  to  the  total  port-opening  -f- 
BC,  or  i-jig-  +  ITV  =  21  inches. 

21.  E'H1 '.      In  a  similar  way  this  is  made  equal  to  the 
total  port-opening  +  B'C',  which  gives  i^  +  •£  =  i-j-^. 

22.  Width  of  Bridge. — This   is  made   equal  to  the  half- 
travel — (EH  or  E'Hf) — inside  lap  -)-  J  inch,  for  a  'minimum. 


DESIGN  OF  AN  ALLEN  OR   "TRICK"    VALVE.  89 

EH  or  E'H'  is  used,  according  to  which  gives  the  greater 
value.  The  inside  lap  to  be  used  is  determined  in  the  same 
way.  E'H'  is  the  value  for  this  case,  and  the  inside  lap  is 
zero  at  both  ends,  so  that  the  rule  gives  2j  —  iJ-J  -|-  J  —  || 
inches.  This  is  the  minimum  thickness  of  the  bridge,  and  if 
it  is  any  greater  than  the  thickness  of  the  rest  of  the  casting, 
the  bridge  must  be  thickened  to  the  required  amount.  If  not, 
the  bridge  may  be  made  of  the  same  thickness  as  the  rest  of 
the  cylinder-casting.  This  will  be  done  in  this  case,  and  the 
bridges  HL  and  H'L'  will  each  be  made  one  inch. 

23.  Width  of  Exhaust-port. — This  is  made  at  least  equal, 
to  the  half-travel  -\-  width  of  steam  port  —  bridge  -f-  inside 
lap;  and  the  width  will  then  be,  in  this  case,  at  least  2j  -["  JyV 
—  i  -j-  o  =  2T5ff ;  and  it  may  be  anything  greater. 

All  the  dimensions  of  the  valve,  valve-seat,  and  false  seat 
are  now  determined,  and  the  valve  may  be  laid  out  in  its 
middle  position,  as  shown  in  Fig.  55,  beginning  at  either  end. 


CHAPTER    X. 
DESIGN   OF  A   DOUBLE-PORTED  VALVE. 

THE  valve  shown  in  Figs.  59-63,  known  as  the  "  double- 
ported  "  valve,  is  another  type  of  valve  used  to  replace  a 
D  valve  when  it  is  desired  to  secure  an  early  cut-off;  and  it 
does  this  in  the  same  general  manner  as  the  Allen  valve,  by 
securing  the  same  port-opening  as  a  D-valve  with  one-half  the 
travel,  or  double  the  opening  with  the  same  travel.  This- 
enables  the  same  point  of  cut-off  to  be  secured  with  a  smaller 
amount  of  outside  lap.  The  lap  and  travel  being  reduced,  it 
follows  that  the  friction  of  the  valve  is  lessened.  From  Figs. 
60  and  61  it  will  be  seen  that  the  steam  enters  the  cylinder 
beneath  the  outer  edges  of  the  valve,  and  that  the  action  of 
the  outer  shell  is  therefore  similar  to  that  of  a  plain  D-valve. 
In  addition  to  this  means  of  admission,  passages  5  and  S'  are 
provided  which  run  all  the  way  through  the  valve,  and  arc 
therefore  open  to  live  steam  from  the  steam-chest  at  all  times. 
These  passages  have  ports  in  the  bottom,  as  shown  at  GH 
and  G'H'.  There  are  corresponding  ports  IL  and  I ' L'  in  the 
valve-seat,  and  steam  is  thus  admitted  through  these  secondary 
ports  as  shown  by  the  arrows  in  Figs.  60  and  61.  If  the 
valve  as  rhown  in  Fig.  60  is  in  its  extreme  left-hand  position, 
the  port  opening  to  steam  is  the  sum  of  G'H'  and  A'C' . 
The  opening  A'C1  is  that  which  would  be  obtained  by  a  plain 
D-valve,  and  therefore,  if  the  valve  is  so  constructed  that 
A ' C'  and  G'H'  are  equal,  the  total  port-opening  is  double  that 
which  would  be  obtained  by  a  plain  slide-valve. 

The  exhaust  passes  through  the  passage  Xy  as  shown  by 

90 


DESIGN   OF  A     DOUBLE-PORTED    VALVE.  9 1 


92  SLIDE-VALVES. 

the  arrows;  and  this  passage  gives  the  valve  a  certain  simi- 
larity to  the  Allen  valve;  but  here  the  similarity  ends,  for  it 
will  be  remembered  that  with  the  Allen  valve  the  passage  is 
used  for  the  admission  alone,  while  here  it  is  only  for  the 
exhaust.  . 

The  action  of  this  valve  is  therefore  the  same  as  would  be 
obtained  by  two  slide-valves,  one  within  the  other.  The 
difference  in  construction  is  that  the  exhaust-passage,  X" ,  of 
the  inner  valve  is  thrown  open  to  the  exhaust-passage,  X,  of 
the  outer  valve.  It  will  be  noticed  that  5  and  S'  are  used 
for  admission  only.  They  are  not  of  the  same  size  through- 
out, but  are  greater  at  the  sides,  where  they  open  to  the 
steam-chest,  as  shown  clearly  in  Fig.  63,  in  which  the  left  half 


FIG'.  63 


of  the  figure  shows  a  transverse  section  through  the  middle  of 
the  valve,  and  the  right-hand  half  is  a  section  through  the 
center  of  the  passage  5.  This  figure  also  shows  the  general 
shape  of  the  passage  X,  which  is  divided  into  two  parts  by  the 
central  web.  Fig.  62  shows  a  section  through  the  center  of 
the  valve,  Figs.  59,  60,  and  61  being  some  distance  to  one 
side. 

This  type  of  valve  is  more  often  applied  to  vertical  engines 
than  to  any  others,  and  is  designed  to  give  equal  cut-offs,  for 
the  following  reasons.  With  equal  cut-offs  secured  by  equal 
laps  the  leads  are  unequal,  and  the  greatest  lead  is  at  the 
crank-end  of  the  cylinder,  as  shown  by  the  diagrams  in 
Chapter  III.  The  crank-end  of  the  cylinder  being  the  lower 


DESIGN   OF  A    DOUBLE-PORTED    VALVE.  93 

end,  the  extra  amount  of  lead  is  useful  in  cushioning  the 
piston  and  piston-rod,  etc.,  which  must,  from  the  nature  of 
the  construction,  have  a  greater  force  on  the  down  than  on  the 
up  stroke.  Another  reason,  or,  rather,  one  in  the  same  line, 
for  not  designing  the  valve  to  have  equal  leads,  is  that  that 
would  give  a  longer  cut-off  on  the  crank-end,  which  is  the  end 
where  the  weight  of  the  moving  parts  tends  to  increase  their 
motion  rather  than  retard  it. 

This  valve,  unlike  the  Allen,  cannot  be  used  on  the  same 
valve-seat  as  a  plain  D  valve  by  the  interposition  of  a  false 
seat,  but  requires  the  valve-seat  to  be  laid  out  with  reference 
to  the  double-ported  valve.  It  is  therefore  impossible  to  take 
off  a  D-valve  from  an  engine  and  put  a  double-ported  valve 
in  place. 

To  show  the  process  of  designing  a  valve,  an  actual  case 
will  be  worked  out. 

PROBLEM. 

GIVEN.  A  cylinder  27  X  24  inches;  ports  24  inches  long; 
revolutions  per  minute,  140;  width  of  bridge,  if  inches; 
inside  lap,  both  ends,  o;  cut-off,  both  ends,  £  stroke;  con- 
necting-rod, 72  inches;  velocity  of  admission,  6000  feet  per 
minute;  velocity  of  exhaust,  4000  feet  per  minute;  lead, 
head  end,  TV  inch. 

REQUIRED.  Outside  lap,  both  ends;  travel  of  valve;  lead, 
crank-end;  maximum  port-opening,  both  ends;  and  to  lay 
out  the  valve  in  its  middle  position  as  shown  in  Fig.  59. 


ORDER   IN   WHICH    THE   DIMENSIONS   ARE   DETERMINED. 


I.  Piston  Speed. — This  is 

140  X  2  X  --  =  $60  feet  per  minute. 


94  SLIDE-VALVES. 

2.  Area  of  Steam- Port. — The  nearest  piston  speed,  to  the 
given  560,  in  Table  IV  is  550,  and  the  port  area  is  given  as 
.138  times  the  piston  area  for  a  velocity  of  4000  feet  per 
minute.     The    piston  area  from   Table  V  is   572.56  square 
inches,  so  that  the  port  area  is 

572.56  X  .138  =  79-01- 

3.  Width  of  Steam-Port. — The  ports  being  24  inches  long, 
the  width  must  be 

79.01  -7-24=  3.3. 

This  is,  however,  the  total  width  of  port  at  one  end  of  the 
cylinder,  and  will  be  taken  as  3^  inches.  There  are  two  ports 
at  each  end  of  the  cylinder,  so  that  the  width  of  each  one  is 
made  3^  -r-  2  =  if  inches.  This  determines  I'L',  C'F'  at  the 
head-end,  and  IL  and  CF  at  the  crank-end,  each  being  made 
if  inches.  The  width  of  the  main  port — that  is,  after  the 
two  ports  join — must  be  made  3^  inches. 

4.  Maximum  Port -Opening  on  the  Head-End. — This  is  the 
opening   which    will    give    the    entering  steam  the  required 
velocity,  and  is  found  as  follows: 

velocity  of  exhaust 

Port-opening  =  port  width  X  — \ — rr~  —t — j — : — : — . 

velocity  of  admission 

This  gives: 

"Pi-M-4-_rvt-»«=»riinrr  —     T  •§•   V 

6000 


.  4000 

Port-opening  =  if  X  %-  -  =  i- 


This  is  the  opening  of  each  port,  and  the  total  opening  at  one 
end  of  the  cylinder  is  therefore  2  X  IT\  =  2f  • 

The  same  result  would  be  obtained  by  using  Table  IV  to 
find  the  area  required  for  free  admission. 

5 .  Valve-  Travel,  Inside  and  Outside  Laps  on  Both  Ends,  and 
Maximum  Port-opening  and  Lead  on  Crank- End. — These  are 
determined  precisely  as  for  the  Allen  valve,  and  the  explana- 


DESIGN   OF  A    DOUBLE-PORTED    VALVE.  95 

tion  given  of  Fig.  57  will  answer  for  this,  the  numerical  values 
only  being  changed.     The  results  obtained  are  as  follows: 

Outside  lap,  head  end \\  inches. 

Outside  lap,  crank  end i         " 

Lead,  crank  end f        "    , 

Maximum  port-opening,  crank  end  ....    i-J-      " 
Valve  travel 5        " 

6.  Width  of  Bridge. — The  minimum  width  allowable  is 
Half-travel  —  width  of  one  port  —  inside  lap  -f-  J  inch. 

2\  if  o          +  J  =  i  inch. 

This  is  less  than  the  if  inches  given  in  the  problem. 
This  rule  is  only  applied  to  see  whether  the  width  given  is 
sufficient  to  prevent  blowing  through.  The  result  obtained 
shows  that  the  if  inches,  which  is  the  thickness  of  the 
cylinder-casting,  is  ample. 

7.  Width  of  GH. — This  is  made  equal  to  the  port-opening 
for  that  end  of  the  cylinder,  and  is  therefore  i£  inches.      GH 
is  at  the  crank  end. 

8.  Width  of  G'H' . — Make  this  equal  to  the  port-opening 
at  its  end  of  the  cylinder,  or  IT3^  inches,  G'H'  being  at  the 
head  end. 

9.  Width  EG. — This  may  be  made  any  convenient  figure, 
and  in  this  case  it  will  be  one  inch. 

10.  Width  E'G'.~  This  is  made  equal  to  EG. 

11.  Width  of  Exhaust-Port. — This  must  be  made  great 
enough   so  that  when   the  valve  is  in   either  of  its  extreme 
positions,  as  shown  in  Figs.  60  and  61,  the  opening  to  exhaust 
is  so  great  that  the  outflowing  steam  will  not  have  a  greater 
velocity  than  4000  feet  per  minute.     The  total  width  of  steam- 
ports  at  one  end  is  that  found  to  be  necessary  for  this  velocity, 
and,  as  the  exhaust-port  takes  care  of  the  exhaust  from  both 


9  SLIDE-  VA  L  VES. 

of  these  ports,  the  free  opening  of  the  exhaust  MK1  ',  Fig.  60, 
or  KM',  Fig.  61,  must  be  equal  to  the  total  width  of  the 
ports  at  one  end  of  the  cylinder.  The  exhaust-port,  MM',  is, 
then  made  of  a  width  equal  to 

Half-travel  +  inside  lap  —  bridge  -\-  port  width  at  one  end. 
2£          +        o         —      if      +        3i  =  4i  inches. 

12.  Width  FI.  —  The  edge  E  must  not  travel  beyond  F  if 
it  is  desired  to  have  a  full  opening  during  exhaust.      That 
gives  the  value  of  FI  as 

Outside  lap,  crank-end  -f-  GH  -\-  EG  -\-  half-travel. 

I  +     i  £  +     I     +     2j  =:  6  inches. 

13.  Width  FT  .  —  This  is  obtained  in  a  similar  manner  to 
FI\  or  it  is  equal  to 

Outside  lap,  head-end  +  G'  H  '  +  E'G'  +  half-travel. 
it  +   i  A    +     i     +        2*  = 


The  edge  E  may  be  allowed  to  overtravel  very  slightly 
the  point  F,  in  case  it  is  an  object  to  shorten  the  valve. 
Doing  so  prevents  the  exhaust  from  being  wide  open  all  the 
time,  but  the  choking  only  occurs  during  a  small  part  of  the 
stroke.  When  this  overtravel  is  allowed,  FI  and  FT  are  to 
be  shortened  by  the  amount  allowed.  No  allowance  is  made 
in  this  case. 

14.  Width  DE.  —  This  is  made  equal  to 

Half-travel,  —  inside   lap,  crank-end  —  amount  that  E  overtravels   F. 
2j  o  0=2%  inches. 

15.  Width  D'E'.  —  This  is  made  equal  to 

Half-travel  —  inside  lap,  head  end  —  amount  E'  overtravels^. 
2^—0  ^0=2%  inches. 


DESIGN  OF  A    DOUBLE-PORTED    VALVE.  97 

The  inside  laps  being  equal,  it  happens  that  DE  and  D'E' 
are  equal.  If  the  inside  laps  are  unequal,  the  values  obtained 
by  the  rules  would  be  unequal;  but  both  DE  and  D'E'  would 
be  made  equal  to  the  larger  value  obtained,  keeping  the  valve 
symmetrical  in  this  respect. 

16.  Width  BC. — This  must  be  so  long  that  D  will  not 
overtravel  B;  for  if  it  did,  the  steam-chest  would  be  in  direct 
communication  with  the  exhaust-passage  X,  allowing  steam 
to  blow  straight  through  the  valve.      Its  width  must  therefore 
be  at  least  equal  to 

Half-travel  —  width  of  one  port  —  inside  lap,  crank-end  +  J  inch. 
2\  if  O  -f-  4  —   l  inch- 

17.  Width  B'C . — This  is  determined,  from  the  same  con- 
siderations that  govern  the  dimension  BC,  to  be 

Half-travel  —  width  of  one  port  —  inside  lap,  head-end  +  \  inch, 
2^  if  —       o  +  J  =  I  inch. 

The  next  thing  in  order  is 


TO    LAY   OUT   THE   VALVE. 

1 8.  Tabulate  the  Dimensions. — The  dimensions  of  the  valve 
and  seat  should  be  tabulated  separately,  in  the  order  in  which 
they  are  to  be  laid  down,  beginning  at  either  end.  In  this 
case  the  start  will  be  made  from  the  crank  end.  The  letters 
apply  to  Fig.  59. 

Valve  Seat. 

BC.  I      inch obtained  from  15 

CF.  if      "     "          "        3 

FI.  6        "     "          "      ii 

IL.  if      "     "          "        3 

LM.  if  "                                  "          "       6 


SLID  E-  VA  L  VES. 

MM'.    4^    inch  .......  .  ----  obtained  from  10 

M'L'.     if      "     ............         "  "  6 

LT.     if      "     ............         "  "  3 

I'F'.     5H    "    ............         "  "  l2 

F'C.     if      "    ............         "  "  3 

B'C.     i        "     ............         "  "  16 

Lay  this  out  on  any  convenient  scale. 


The  Valve. 

AC.  I     inch  outside  lap  ........  obtained  from  5 

DF.  o  "      inside  lap     .  .......  given 

DE.  2\  "  .......  obtained     "    13 

EG.  i  "   •  ........        "  "      8 

GH.  i  £  "  ........        "  <4      6 

J/A'.  i  «      outside  lap    .......         "  "      5 

LK.  o  "      inside  lap    ......  .  given 

K'L'.  o  "         "       "       ........      " 

I'H'.  ij  "      outside  lap  .....  ...  obtained  from  5 

H'G'.  IT875-  "      port  opening  ......        "  "      7 

G  '  E'  '  .  i  "  ........  obtained  from  9 

E'D'.  2%  *>  ........        "  "    14 

F'D'.  o  "      inside  lap    .......  ,  given 

ij  "      outside  lap  ........  obtained  from  5 


Reference  to  Fig.  63  will  show  that  the  exhaust-passage 
XX,  Fig.  59,  is  divided  into  two  portions,  X'  and  X"  ,  when 
it  passes  over  either  of  the  steam-passages,  as  5.  Care  must 
be  taken  to  make  the  valve  of  so  great  a  height  that  the  sum 
of  the  areas  of  these  two  portions  shall  be  at  least  as  great  as 
the  total  area  of  the  ports  at  one  end  of  the  cylinder  in  order 
to  maintain  the  proper  velocity  of  exhaust. 


CHAPTER   XL 
VALVE-SETTING. 

HAVING  designed  a  valve  and  having  it  in  place  on  the 
engine,  the  next  thing  to  be  considered  is  the  proper  method 
of  setting  it  in  order  to  secure  the  best  results.  The  method 
to  be  employed  will  depend  on  the  design  of  the  valve— 
whether  it  is  intended  to  secure  equal  cut-offs  on  the  two 
ends,  or  whether  it  was  made  to  give  equal  leads;  or,  again, 
whether  the  cut-offs  and  leads  were  equalized,  according  to 
the  method  given  in  Chapter  VI. 

In  any  case  the  first  step  is  to  put  the  engine  on  the  center, 
by  which  is  meant  that  the  piston  is  at  the  end  of  its  stroke 
and  the  crank  and  connecting-rod  are  in  the  same  straight 
line.  This  operation  must  be  performed  very  precisely,  as 
little  variation  in  the  position  of  the  crank  from  the  line  of 
stroke  or  piston  from  the  end  of  the  stroke  will  make  a  large 
difference  in  the  position  of  the  valve.  This  is  because  the 
eccentric  is  at  or  near  its  middle  position  at  the  time  the  crank 
is  on  the  center.  The  crank-pin  is  moving  vertically  while 
the  eccentric  is  moving  horizontally  at  that  instant,  so  that  a 
small  angular  movement  of  the  eccentric  will  make  a  great 
difference  in  the  position  of  the  valve.  The  crank,  however, 
is  moving  vertically,  and  a  comparatively  large  angular  move- 
ment of  that  will  only  move  the  cross-head  a  little  ways  from 
the  end  of  its  travel.  Consequently  the  dead-center  must 
be  located  very  exactly- or  the  valve-setting  will  be  thrown 
out. 

99 


IOO 


SLIDE-VALVES. 


TO   PUT   AN    ENGINE    ON   THE    CENTER 

The  engine  is  put  on  the  center  by  moving  the  cross-head 
a  measured  distance  on  each  side  of  its  extreme  travel  and 


FIG.  64 

measuring  the  amount  of  the  movement  of  the  fly-wheel  by 
means  of  marks  on  the  rim.     This  distance  is  bisected,  and 


FIG.  65 

the  middle  point  determines  the  true  center.     The  practical 
method  of  doing  this  is  shown  in  Figs.  64,  65,  and  66. 

First.  Turn  the  engine  in  the  direction  in  which  it  is  to 
run  until  the  cross-head  is  nearly  at  the  end  of  its  stroke,  as 
shown  in  Fig.  64. 


VA  L  VE-SE  TTING. 


101 


Second.  With  the  cross-head  in  this  position,  take  a  piece 
of  chalk  and  make  a  mark  across  the  cross-head  and  guides,  as 


Fief.  66 


shown  at  A,  Fig.  64.  This  mark  may  be  put  anywhere  on 
the  cross-head,  as  it  serves  only  as  a  reference-mark. 

Third.  Fix  an  upright  pointer  as  near  as  possible  to  the 
face  of  the  fly-wheel,  as  shown  at  B,  and  mark  the  height 
reached  by  its  end,  as  shown  at  C. 

Fourth.  Turn  the  engine,  still  in  the  direction  in  which  it 
is  to  run,  until  the  mark  on  the  cross-head  again  comes  even 
with  the  mark  on  the  guides,  as  shown  in  Fig.  65. 

Fifth.  Make  another  chalk-mark  on  the  fly-wheel  opposite 
the  end  of  the  upright,  as  shown  at  C' ,  Fig.  65.  The  posi- 
tion of  the  first  mark  after  this  movement  is  then  as  shown 
at  C.  The  distance  CC'  then  represents  the  amount  that  the 
fly-wheel  has  moved,  while  the  cross-head  has  moved  from  A 
out  to  the  dead-center  and  back  again  to  A.  That  is,  the 
cross-head  movement  has  been  twice  the  distance  from  A  to 
the  dead-center.  Consequently,  if  the  wheel  revolved  until 
the  pointer  comes  midway  between  C  and  Cf,  the  engine  will 
be  on  the  dead-center. 

Sixth.  Find  the  point  midway  between  C  and  C' .  This 
is  D,  Fig.  65. 


-1 02  SLID  E-  VA  L  VES. 

Seventh.  Turn  the  engine,  still  in  the  direction  in  which  it 
is  to  run,  until  the  mark  D  comes  opposite  the  end  of  the 
pointer,  as  shown  in  Fig.  66.  The  engine  is  then  on  the 
dead-center. 

Eighth.  With  the  engine  on  the  dead-center,  make  another 
chalk-mark  on  the  guides  opposite  the  one  on  the  cross-head, 
as  at  E,  Fig.  66.  This  is  to  serve  as  a  reference-mark,  and 
the  first  mark  is  then  to  be  erased. 

The  dead-center  at  the  other  end  is  found  in  the  same 
manner. 

When  putting  the  engine  on  the  center,  or  when  perform- 
ing any  of  the  other  operations  of  valve-setting,  the  engine 
must  always  be  turned  in  the  direction  in  which  it  is  to  run. 
This  is  because  any  lost  motion,  back-lash,  or  play  in  the 
moving  parts  will  then  affect  the  valve  in  setting  precisely  as 
under  running  conditions.  The  wheel  must  never  be  turned 
beyond  the  required  point  and  then  back  to  it,  as  the  lost 
motion  would  allow  a  considerable  movement  of  the  fly-wheel 
to  take  place  without  a  corresponding  motion  of  the  cross- 
head  ;  and  the  valve-setting  would  then  be  thrown  put  of  true. 

TO    SET   A   VALVE   FOR   EQUAL   LEADS. 

First  Method. 

First.  Put  the  engine  on  the  dead-center  at  one  end  of  its 
stroke,  using  the  method  just  described. 

Second.  Give  the  eccentric  as  nearly  as  possible  the 
proper  amount  of  angular  advance,  as  determined  from  the 
valve  diagrams,  taking  care  that  the  amount  given  shall  be 
more,  rather  than  less,  than  the  required  amount. 

Third.  Adjust  the  length  of  the  valve-stem  or  eccentric- 
rod  until  the  lead  at  the  end  of  the  stroke  for  which  the 
adjustment  is  then  being  made  is  equal  to  the  required 
amount. 

Fourth.   Turn  the  engine  to  the  other  dead-center. 


VA  L  VE-SE  T  TING.  1 03 

Fifth.  Measure  the  lead  at  that  end.  If  the  measure- 
ment is  the  same  as  that  at  the  first  end,  the  eccentric  is  in  its 
proper  place,  aad  it  only  remains  to  secure  it  there;  but  the 
chances  are  that  the  leads  will  be  unequal,  and  in  that  case 
the  next  step  is  the 

Sixth.  Correct  half  the  difference  in  the  leads  by  changing 
the  length  of  the  valve-stem. 

Seventh.  Correct  the  remaining  half  of  the  difference  by 
moving  the  eccentric.  It  will  be  apparent  at  once,  from  the 
nature  of  the  difference,  in  which  direction  the  eccentric  is  to 
be  moved. 

Eiglith.  Turn  back  to  the  other  center  and  measure  the 
lead.  If  it  is  the  required  amount,  the  setting  is  complete. 
If  not,  repeat  operations  Sixth  and  Seventh  until  the  leads  are 
equal. 

Ninth.   Secure  the  eccentric  in  place. 

When  a  valve-gear  has  a  rocker,  the  latter  is  usually  de- 
signed to  swing  to  an  equal  angle  on  each  side  of  the  perpen- 
dicular, and  in  any  case  the  length  of  the  valve-stem  must  be 
such  that  the  rocker  will  move  as  designed.  This  being  so, 
it  is  evident  that  in  performing  the  sixth  and  seventh  opera- 
tions— those  of  correcting  the  variation  in  leads  at  the  two 
ends  of  the  cylinder — the  length  of  the  valve-stem  must  be 
changed  very  little,  if  any,  making  the  sixth  operation  very 
small,  and  the  greater  part  of  the  adjustment  must  be  made 
in  the  seventh  operation,  that  of  changing  the  position  of  the 
eccentric. 

Second  Method. 

This  method  is  a  convenient  one  when  it  is  difficult  to 
turn  tlu  engine  over.  It  is  applicable  only  to  that  class  of 
valves  having  harmonic  motion.  Harmonic  motion  is  such 
as  that  of  the  foot  of  a  perpendicular  from  the  crank-pin  upon 
the  line  of  stroke  when  the  motion  of  the  crank-pin  is  uniform. 
This  is  the  case  in  an  ordinary  engine,  and  an  ordinary  slide- 


1 04  SLID  E-  VA  L  VES. 

valve  has  harmonic  motion.  Among  the  valve-gears  in  which 
the  valve  does  not  have  harmonic  motion  may  be  mentioned 
a  slide-valve  having  equal  lead  and  the  cut-offs  equalized  by 
means  of  a  rocker  or  bell-crank  lever,  and  the  link  motion  and 
radial  gears. 

When  the  motion  is  harmonic,  the  maximum  port-open- 
ings will  be  equal  when  the  leads  are  equal. 

First.   Loosen  the  eccentric  on  the  shaft. 

Second.  Turn  the  eccentric  until  it  gives  the  maximum 
port-opening,  first  at  one  end  and  then  at  the  other. 

Third.  If  the  maximum  port-openings  are  not  equal — and 
the  chances  are  that  this  will  be  the  case — make  them  so,  by 
changing  the  length  of  the  valve-stem  by  half  the  difference, 
thus  adjusting  the  length  of  the  valve-stem. 

Fourth.  Put  the  engine  on  the  center.  This  is  the  only 
time  that  it  is  necessary  to  perform  this  operation. 

Fifth.  Turn  the  eccentric  to  give  the  proper  lead,  thus 
adjusting  the  angle  of  advance. 

Sixth.   Secure  the  eccentric  in  place.* 

TO    SET  A  VALVE   FOR   EQUAL   CUT-OFFS. 

First.  Put  the  engine  on  one  dead-center — say  the  head- 
end. 

Second.  Give  the  eccentric,  as  nearly  as  can  be  judged, 
the  angle  of  advance  determined  by  the  valve-diagram;  tak- 
ing care  that  if  there  is  any  difference,  it  shall  be  in  excess  of 
the  proper  amount  rather  than  less. 

Third.  Give  the  valve  the  correct  amount  of  lead,  as 
nearly  as  possible. 

Fourth.  Move  the  engine  in  the  direction  in  which  it  is  to 
run  until  cut-off  occurs. 

Fifth.  Measure  the  distance  that  the  cross-head  has  moved, 

*The  author  wishes  to  acknowledge  his   indedbteness  to  Peabody  on 
Valve-gears  for  this  method. 


VAL  VE-SETTING.  10$ 

up  to  this  point,  from  the  end  of  the  stroke.     This  is  found 
by  means  of  the  chalk-marks  on  the  guides  and  cross-head. 

Sixth.  Turn  the  engine  in  the  direction  in  which  it  is  to 
run  until  cut-off  occurs  on  the  return  stroke. 

Seventh.  Measure  the  travel  of  the  cross-head  from  the 
beginning  of  the  return  stroke  up  to  cut-off.  If  this  is  the 
same  as  on  the  forward  stroke,  the  valve  is  set  correctly,  and 
the  eccentric  should  then  be  secured  in  place.  But  it  is 
hardly  probable  that  this  result  will  be  secured  on  the  first 
trial;  and  in  that  case  the  next  operation  is  the 

Eighth.  Correct  the  difference  in  cut-offs  by  changing  the 
length  of  the  valve-stem.  If  the  cut-off  is  earlier  on  the 
crank-end  or  return  stroke,  the  valve-stem  should  be  length- 
ened. If  it  is  earlier  on  the  head-end  or  forward  stroke,  the 
valve-stem  should  be  shortened. 

Ninth.  Put  the  engine  on  the  head-end  center  again,  and 
adjust  the  lead  by  moving  the  eccentric. 

Tenth.  Test  the  cut-offs  and  see  whether  or  not  they  are 
equal.  If  they  are,  the  adjustment  is  finished.  If  not,  repeat 
the  eighth  operation  until  they  are  equal,  and  then — 

Eleventh.   Secure  the  eccentric  in  place. 

As  pointed  out  in  Chapter  III,  designing  a  valve  for  equal 
cut-offs  will  make  the  leads  unequal.  At  the  time  the  valve 
is  designed  the  lead  can  be  measured  at  each  end  of  the 
cylinder,  and  the*  operation  of  setting  the  valve  can  be  per- 
formed by  using  the  first  method  given  above  for  equal  leads, 
except  that  the  leads,  instead  of  being  made  equal,  are  made 
equal  to  the  required  amounts.  In  addition,  the  travel  of  the 
cross-head  from  the  beginning  of  the  stroke  up  to  cut-off 
must  be  determined  and  the  setting  completed  by  the  eighth 
and  ninth  operations  just  given. 

.  By  means  of  any  of  these  methods,  it  is  assured  that  the 
action  of  the  valve  shall  be  just  as  intended  at  admission  or 
cut-off.  This  also  assures  the  fact  that  any  irregularity  of 
action,  or  error  of  design,  caused  by  neglecting  the  angularity 


io6 


SLIDE-VALVES. 


of  the  eccentric-rod,  will  not  affect  the  valve  while  either 
opening  or  closing,  but  will  make  itself  felt  while  the  port  is 
either  opened  or  closed;  which  is  of  no  particular  conse- 
quence. 

Now,  having  set  the  valve,  it  is  of  importance  to  make  a 
distinct  set  of  reference-marks  on  the  eccentric,  so  that  it  will 
be  possible  to  set  it  in  place  again  very  readily  if  it  slips,  thus 
avoiding  vexatious  delays,  or,  perhaps,  the  necessity  of  going 
over  the  whole  ground  cf  valve-setting  again. 

When  any  one  of  the  preceding  methods  of  valve-setting 
is  employed,  it  is  necessary  to  remove  the  cover  of  the  steam- 
chest  in  order  to  observe  the  action  of  the  valve,  measure 
leads,  etc. 

TO    SET  A   VALVE   WITH   THE    CHEST-COVER   ON. 

After  having  once  set  the  valve  for  equal  leads,  with  the 
cover  off,  it  is  possible  to  arrange  things  so  that  thereafter  the 
valve  can  be  set  with  the  chest-cover  on,  and,  if  absolutely 
necessary,  with  steam  on.  This  method  is  illustrated  in 
Figs.  67  and  68. 

After  having  set  the  valve,  the  engine  is  placed  on  the 
center,  and  a  reference-mark  made  on  the  valve-stem  some- 
where outside  of  the  stuffing-box,  as  shown  at  A,  Fig.  67. 
Then  another  mark  is  made  somewhere  on  the  chest-cover,  as 
at  B.  Next  a  piece  of  heavy  wire  is  taken,  and  a  tram  or 
spanner,  T,  Fig.  67,  is  made,  of  such  length  that  it  will  reach 
from  A  to  B.  Then  the  engine  is  put  on  the  other  center, 
which  operation  will  change  the  distance  of  the  mark  on  the 
valve-stem  from  the  mark  on  the  chest-cover  to  AB,  Fig.  68. 
Another  spanner,  T' ,  is  then  made  which  will  reach  that  dis- 
tance. Then,  when  it  is  again  necessary  to  set  the  valve,  the 
operation  can  be  performed  by  employing  the  first  method  for 
equal  leads,  using  the  trams  to  determine  the  equality  of  the 
leads,  as  follows: 


VAL  VE-SE  TTING. 


107 


First.   Put  the  engine  on  one  center,  say  the  head-end. 

Second.  Give  the  eccentric  the  proper  amount  of  angular 
advance,  as  nearly  as  possible,  making  it  too  great  rather 
than  too  little. 

Third.  Adjust  the  length  of  the  valve-rod  until  the  lead 
at  the  head-end  is  the  proper  amount;  that  is,  until  the  tram 
T  reaches  from  A  to  B. 

Fourth.  Turn  the  engine  to  the  other  center. 

Fifth.  Measure  the  crank-end  lead.  This  is  done  by  using 
the  tram  T' .  Remember  that  when  it  reaches  from  A  to  B, 


FIG.  67 


as  in  Fig.  68,  the  leads  are  equal.  When  it  reaches  to  any 
other  point,  the  lead  is  changed  by  the  distance  from  A  to 
the  point  which  the  tram  marks  on  the  valve-stem.  If  the 
tram  spans  over  to  the  crank  side  of  A,  the  lead  is  too  great 
at  the  crank-end.  If  the  tram  reaches  to  a  point  on  the 
cylinder  side  of  A,  the  lead  at  the  crank-end  is  too  little. 

The  rest  of  the  operations  are  the  same  as  given  for  equal 
leads. 

A  variation  of  this  method  consists  in  using  but  one  tram, 
and  making  two  marks  on  the  valve-stem,  at  the  points 
reached  by  the  tram  with  the  engine  on  the  two  -enters. 
The  method  of  setting  is  then  the  same.  This  is  frequently 
used  on  locomotives. 


CHAPTER    XII. 


SHAFT-GOVERNORS.     GENERAL  PRINCIPLES   AND   TYPES. 

IT  was  shown  in  the  first  chapter  that  in  order  to  reverse 
the  direction  of  rotation  of  an  engine  it  is  necessary  to  move 
the  eccentric  around  the  shaft  past  the  crank  until  it  makes 
the  same  angle  with  it  on  the  opposite  side  that  it  did  in  its 
original  position.  That  is,  if  the  arrangement  of  crank  and 
eccentric  shown  in  Fig.  69  will  cause  the  engine  to  run  over, 


1C 


0' 


o 


1C' 

\ 
\ 
\ 


S 


FIG.  69 


FIG.  70 


as  shown  by  the  arrow,  the  arrangement  shown  in  Fig.  70 
will  reverse  the  engine,  the  angle  C'O'E'  being  equal  to  the 
angle  COE.  With  this  arrangement  no  rocker-arm  is  em- 
ployed between  the  eccentric-rod  and  the  eccentric. 

Figs.  71  and  72  show  the  arrangement  when  a  rocker-arm 


100 


SHA  FT-  G  O  VERNORS. 


I09 


is  employed.  The  angle  CO' E'  is  equal  to  the  angle  COE, 
just  as  before,  and  the  direction  of  rotation  has  been  reversed, 
as  shown  by  the  arrows. 

In  both  cases  the  angle  through  which  the  eccentric  has 
been  turned  to  effect  the  reversal  is  equal  to  180° — twice  the 
angle  of  advance. 


V 


E  — 


FIG.  71 


rr  FIG.  72 


There  are  numerous  means  by  which  the  eccentric  can  be 
moved  around  the  shaft ;  and  they  may  be  divided  into  two 
general  classes: 

First  Class.  Movable  eccentrics  with  which  the  engine 
must  be  stopped  in  order  to  effect  the  reversal. 

Second  Class.  Movable  eccentrics  with  which  the  reversal 
can  be  effected  while  the  engine  is  in  motion. 


FIRST    CLASS. 


This  is  by  far  the  simpler  class,  and,  for  reasons  which 
should  be  very  obvious,  it  is  but  seldom  employed,  although 
it  was  used  on  early  stationary  and  locomotive  engines. 
Figs.  73  and  74  illustrate  an  eccentric  of  this  type.  The  shaft 
carries  a  disk  which  is  forged  or  cast  on  it,  or  fastened  in 


HO  SLIDE-  VA  L  VES. 

place.  This  disk  is  slotted  in  an  arc  of  a  circle  about  the 
center  of  the  shaft  O.  The  eccentric  is  loose  on  the  shaft  and 
is  slotted  in  a  similar  manner.  A  bolt  is  used  to  fasten  the 
two  together.  With  the  eccentric  in  one  extreme  position, 
such  as  the  one  shown  in  Fig.  73,  the  angle  of  advance  is 
XOD.  The  arc  of  the  slot  is  made  of  such  a  length  that 
when  the  eccentric  is  slipped  around  so  that  it  is  bolted  in  its 
other  extreme  position,  as  shown  in  Fig.  74,  the  angle  of 
advance  is  YOD' ',  equal  to  XOD  of  Fig.  73.  In  both  figures 
the  path  of  the  eccentric  center  is  the  dotted  circle  shown. 

That  this  change  in  the  angle  of  advance  will  result  in  a 
reversal  of  the  engine  should  be  understood  from  Chapter  I. 
Figs.  75  and  76,  the  valve  diagrams  corresponding  to  Figs. 
73  and  74  respectively,  will  aid  in  the  comprehension  of  the 
fact  that  the  events  of  the  stroke  occur  at  the  same  points  in 
both  cases.  The  angles  of  advance  are  laid  off  on  opposite 
sides  of  the  vertical  in  the  two  figures  for  the  sake  of  em- 
phasis, but,  as  explained  in  Chapter  III,  this  is  not  necessary. 

Fig.  77  shows  another  type  of  this  class.  The  eccentric- 
slot  is  replaced  by  a  pin  which  projects  through  the  slot  in 
the  disk,  and  is  secured  in  place  by  a  jam-nut  on  the  opposite 
side.  The  same  results  will  be  secured  by  this  type  as  with 
the  one  shown  in  Figs.  73  and  74,  and  the  diagrams  75  and 
76  apply  to  it  as  well. 

SECOND    CLASS. 

This  is  the  more  important  class,  possessing  many  points 
of  advantage  over  the  first  class,  and  is  the  one  which  is 
employed  on  all  or  nearly  all  high-speed  engines,  where  they 
are  used  as  "shaft-governors." 

The  fundamental  principle  in  this  class  is  to  have  the 
eccentric  slotted  to  clear  the  shaft,  and,  by  moving  the  eccen- 
tric across  the  shaft,  change  the  angle  of  advance  as  required. 

Eccentrics  of  this  class  are  of  either  one  of  two  kinds: 


SHA  FT-GO  VERNORS. 


Ill 


112 


SLIDE-VALVES. 


SHA FT-GO  VERNORS.  1 1  $ 

I.  "Swinging"  Eccentrics;  in  which  the  eccentric  is 
pivoted  at  some  point  which  is  secured  to  the  shaft  and  rotates 
with  it.  In  that  case  the  eccentric  swings  across  the  shaft 
in  the  arc  of  a  circle,  and  the  slot  is  therefore  curved. 


2.  "  Shifting"  Eccentrics;  which  move  squarely  across  the 
shaft,  and  the  slot  is  of  course  straight. 

The  first  question  to  be  considered  is  the  effect  of  moving 
the  eccentric  across  the  shaft  and  holding  it  in  any  given 
position;  the  means  by  which  this  movement  is  effected  being 
reserved  for  later  discussion. 


114 


SLIDE-VALVES. 


SWINGING    ECCENTRICS. 

Figs.  78  and  79  show  the  two  extreme  positions  of  an 
eccentric  of  this  class.  It  is  pivoted  at  the  point  P,  which  is 
usually  located  on  the  arm  of  a  small  fly-wheel  called  the 
* 'governor- wheel  "  or  "spider,"  which  is  securely  keyed  to 
the  shaft.  When  in  its  upper  position,  Fig.  78,  the  angle 
between  the  crank  and  the  eccentric  is  COE]  and  when  in  its 
lower  position  it  is  C 'O 'E' ',  Fig.  79.  The  shaded  area  in  each 


figure  is  the  shaft,  and  the  path  of  the  eccentric  center  is 
shown  by  the  dotted  circle. 

The  manner  of  laying  out  an  eccentric  in  this  type  is 
.shown  in  Fig.  80.  The  point  /*,  about  which  the  eccentric 
is  to  swing,  is  generally  fixed,  and  the  center  O  of  the  shaft  is 
also  fixed.  Then  draw  the  line  VZ  through  O  and  C,  and 
through  O  draw  XY  perpendicular  to  VZ.  The  angle  of 
advance  having  been  determined,  lay  off  XOD  equal  to  that 
angle.  Then  describe  the  circle  whose  center  is  O  and  whose 

o 

radius,  OE,  is  equal  to  the  eccentricity.  This  will  cut  the 
line  OD  at  E,  which  is  the  center  of  the  eccentric.  Now 
about  P  as  a  center  describe  an  arc  passing  through  E, 
Again  with  P  as  a  center  describe  an  arc  passing  through  the 
center  of  the  shaft  O.  Now  E' ,  which  is  where  the  eccentric 


SHA  FT-  G  O  VERNORS.  1 1  5 

center  must  be  in  order  to  secure  a  reversal  of  the  engine, 
will  of  course  lie  on  the  arc  through  E.  Then  the  eccentric 
center  must  be  shifted  through  the  arc  EE'  to  secure  the 
reversal.  The  next  thing  is  to  determine  the  length  of  slot 
required  to  allow  this  movement,  and  the  smallest  diameter 
of  eccentric.  Therefore  from  e,  the  point  where  the  arc  EE' 
cuts  the  line  VZ,  lay  off  the  arc  ee'  equal  to  EE' .  Draw  e'P, 
which  cuts  the  arc  drawn-  through  O  at  O'.  Now  the  slot 
must  be  drawn  in  as  shown.  That  is,  the  lower  center  is  O, 
and  the  radius  is  just  enough  larger  than  the  radius  of  the 
shaft  to  permit  it  to  clear.  The  upper  center  is  O' ,  and  the 
radius  is  of  course  the  same  as  the  lower.  The  rest  of  the 
slot  outline  is  made  up  of  two  arcs  having  P  as  their  center. 

It  is  obvious  that  the  center  line  of  the  crank  must  coincide 
with  VZ,  and  the  spider  or  containing-wheel  of  the  governor 
must  be  keyed  to  the  shaft  so  that  this  result  will  be  obtained. 

Now  suppose  the  eccentric  to  have  swung  around  so  far 
that  the  center  is  at  I,  Fig.  80.  What  is  the  result  ?  The 
angular  advance  has  been  increased  to  XO\,  and  the  eccen- 
tricity has  been  decreased  to  Oi.  (In  order  to  have  the 
eccentricity  remain  the  same,  it  would  be  necessary  to  have 
the  eccentric  move  around  O  as  a  center.)  Now,  the  effect 
of  increasing  the  angular  advance  is  to  make  the  events  of  the 
stroke  all  occur  earlier.  Decreasing  the  travel  makes  the 
admission  later,  cut-off  earlier,  release  earlier,  and  compres- 
sion earlier,  as  shown  in  Table  III.  The  combined  effect  of 
the  two  changes  can  best  be  understood  by  constructing  the 
diagram.  Fig.  81  shows  the  diagrams  for  five  positions  of  the 
eccentric  center  in  Fig.  80.  Diagrams  E  and  E'  correspond 
to  the  extreme  positions  of  the  eccentric,  or,  as  they  are  called, 
*' full-gear  forward"  and  "full-gear  backward/'  Diagram  2 
shows  the  steam  distribution  when  the  eccentric  center  is  at 
2,  Fig.  80,  midway  between  its  two  extremes;  that  is,  when 
the  eccentric  is  in  "mid-gear,"  Diagrams  I  and  3  show  the 


SLIDE-VALVES. 

results  obtained  with  the  eccentric  center  at  I  and  3  respect- 
ively, Fig.  80. 

It  will   be   noticed  that  the  cut-off  varies  more  than   the 
admission  or  lead,  because  both  changes — of  angular  advance 


and  eccentricity — affect  the  cut-off  in  the  same  way,  making 
it  earlier;  while  the  admission  is  made  later  by  the  decreased 
travel,  and  earlier  by  the  increased  angular  advance.  These 
two  opposite  changes  in  the  lead  may  neutralize  each  other, 
but  only  when  the  center,  P,  about  which  the  eccentric 
swings  is  removed  to  infinity — that  is,  when  the  arc  through 
which  it  swings  becomes  a  straight  line,  and  the  eccentric 
moves  straight  across  the  shaft.  The  valve  diagrams  show 
that  the  release  varies  less  than  the  compression,  with  tlu 
swinging  eccentric. 


SHAFT-GO  VERNORS. 


117 


SLIDE-VALVES. 


SHIFTING    ECCENTRICS. 

This  is  the  type  in  which  the  lead  is  constant,  as  stated  in 
the  preceding  paragraph.  It  differs  from  the  swinging  eccen- 
tric more  in  degree  than  in  kind.  Figs.  82  and  83  show  one 


Y 

FIG.  82 


of  this  kind  in  its  two  extreme  positions.  It  will  be  noticed 
that  the  cut-off  has  a  greater  range  of  variation  in  this  than 
in  the  swinging  eccentric,  because  the  valve  travel  has  a 
greater  range.  This  will  be  clearly  understood  by  reference 
to  Fig.  80,  where  the  straight  line  from  E  to  E'  shows  the 
path  which  would  be  followed  by  the  center  of  a  shifting 
eccentric  designed  to  secure  the  same  results  as  the  swinging 
eccentric.  The.  eccentricity  being  the  distance  from  the 
center  of  the  shaft  to  the  center  of  the  eccentric,  the  variation 
is  evidently  the  greater  with  the  shifting  eccentric. 

The  next  thing  is  to  demonstrate  that  these  variations  in 
the  events  of  the  stroke  can  be  made  to  serve  a  useful  pur- 
pose. 

Suppose  an  engine  be  running  at  full  load ;  the  eccentric 


SHA  FT-  G  O  VERNORS.  1 1 9 

being  fixed  in  place.  Then  suppose  that  a  large  part  of 
the  load  is  suddenly  thrown  off.  If  the  steam-pressure 
remains  the  same,  the  engine  will  begin  to  speed  up;  and  if 
enough  load  has  been  thrown  off,  the  increase  in  speed  may 
be  sufficient  to  result  in  a  bursting  fly-wheel. 

Now  suppose  that  the  engine  is  fitted  with  a  movable 
eccentric,  and  that  when  the  full  load  is  on  the  eccentric 
is  in  full  gear  forward,  giving  the  latest  admission  and  latest 
cut-off.  Again,  suppose  that  the  load  is  thrown  off,  just  as 
before,  but  at  the  same  time  the  eccentric  is  moved  in  toward 
the1  shaft.  What  will  happen  ?  Under  the  lighter  load  the 
engine  will  tend  to  speed  up,  but  the  earlier  admission,  due 
to  the  increased  angular  advance,  will  cushion  the  piston, 
tending  to  decrease  the  speed  of  the  piston;  and  the  cut-off 
will  come  earlier  in  the  stroke,  also  tending  to  reduce  the 
piston  speed  by  shortening  the  length  of  time  the  moving 
force  is  applied.  In  other  words,  the  period  of  admission, 
while  remaining  the  same,  is  divided  more  nearly  evenly 
between  the  two  strokes. 

Now,  with  the  cut-off  made  earlier,  the  engine  will  not 
develop  so  much  power;  and  the  change  may  be  just  suffi- 
cient to  adapt  the  engine  to  the  lighter  load.  A  light  load 
at  given  speed  requires  less  power  than  a  heavy  load  at  the 
same  speed. 

Figs.  84  to  87,  inclusive,  illustrate  various  forms  of  shaft- 
governors.  The  general  principle  of  their  construction  is  as 
follows : 

A  small  wheel  or  spider  is  keyed  on  the  shaft,  as  stated 
in  the  previous  chapter.  This  wheel  carries  weights  which 
are  pivoted  on  the  arms  or  ring,  so  that  under  the  influence 
of  centrifugal  force  they  fly  outward,  this  motion  being 
opposed  by  springs.  The  weights  are  linked  to  the  eccen- 
tric, so  that  a  motion  of  the  weights  will  cause  a  correspond- 
ing motion  of  the  eccentric.  Now,  if  the  engine  speed 
remains  constant,  the  centrifugal  force  remains  constant,  and 


120 


SLIDE-VALVES. 


FIG.  84 


FIG.  85 


SHAFT-GO  VERNORS. 


121 


the  springs  will  be  stretched  a  certain  amount,  and  the 
eccentric  will  be  held  in  one  position.  If,  now,  the  load  be 
lessened,  the  engine  will  speed  up  momentarily,  increasing 


FIG.  86 


FIG.  87 

the  centrifugal  force,  thus  throwing  the  weights  outward  and 
moving  the  eccentric  across  the  shaft,  shortening  the  cut-off 
and  adapting  the  power  to  the  load.  If  the  load  is  increased, 


122  SLIDE-VAL  VES. 

the  engine  will  slow  down,  decreasing  the  centrifugal  force  so 
that  the  springs  will  pull  the  eccentric  back,  lengthening  the 
cut-off.  .  In  this  way  the  events  of  the  stroke  are  regulated  so 
as  to  keep  the  speed  of  rotation  sensibly  constant  at  all  loads. 

Figs.  84  and  85  show  two  positions  of  the  Westinghouse 
governor,  Fig.  84  representing  the  governor  when  at  rest  or 
running  below  its  normal  speed;  that  is,  at  its  latest  cut-off. 
Fig.  85  shows  the  governor  when  in  the  position  correspond- 
ing to  its  earliest  cut-off;  that  is,  when  the  engine  has 
speeded  up  considerably  above  the  desired  point  and  the 
governor  is  acting  to  bring  it  down  to  the  proper  point. 

The  governor-weights  B  and  B  are  pivoted  to  the  wheel 
A,  which  revolves  with  the  shaft  M  at  b  and  b.  The  upper 
weight  carries  a  link,  PT,  one  end  of  which,  P,  is  pivoted  to 
the  governor-weight,  and  the  other  end,  7",  is  pivoted  to  the 
eccentric  casting.  The  weights  are  connected  by  the  link  ee. 
The  springs  D  and  D  are  fastened  to  the  weights  and  to  the 
wheel  or  disk,  as  shown;  thus  resisting  any  tendency  of  the 
weights  to  fly  outward.  Under  an  increase  of  speed  the 
centrifugal  force  increases,  and  the  springs  are  extended,  thus 
allowing  the  eccentric  to  move  across  the  shaft. 

Fig.  86  shows  the  Straight-Line  governor,  which  is  con- 
tained within  the  fly-wheel.  The  center  of  the  shaft  is  at  O. 
One  of  the  fly-wheel  arms,  N,  has  a  pivot  on  which  the 
weight-lever,  WNM,  is  hung.  The  end,  M,  of  the  lever  is 
connected  to  the  eccentric  casting  at  Fand  to  the  spring  at 
Z-,  by  the  link  MVL.  The  spring  is  secured  to  a  boss,  (9,  on 
the  fly-wheel  rim,  as  shown,  and  the  eccentric  is  of  the  swing- 
ing type,  being  pivoted  at  5.  The  whole  arrrangement 
occupies  the  position  shown  when  the  engine  is  at  rest;  it 
will  be  noticed  that  the  eccentric  is  in  its  extreme  position, 
thus  securing  the  greatest  cut-off,  as  required  when  the  engine 
starts  up.  When  the  engine  starts  up  the  action  of  the  cen- 
trifugal force  will  force  the  weight  outward,  thus  moving  the 
eccentric  across  the  shaft. 


Fig.  87  represents  the  Buckeye  governor.  When  the 
engine  is  at  rest  the  springs  F F  hold  the  weights  A  A  against 
the  inner  stops.  When  the  engine  starts  up  the  weights  tend 
to  fly  outward ;  and  when  a  certain  rotative  speed  is  reached 
they  move  away  from  the  stops,  thus  stretching  the  springs. 
The  eccentric  being  connected  to  the  weights  by  the  rods  BB, 
it  is  moved  around  the  shaft,  thus  causing  an  earlier  cut-off. 
When  a  point  of  cut-off  corresponding  to  the  load  has  been 
established,  the  speed  ceases  to  increase,  and  the  spring  pull 
and  the  centrifugal  force  will  balance  each  other  as  long  as 
there  are  no  changes  in  the  load  on  the  engine.  If  the  load 
is  increased,  or  the  pressure  of  the  steam  is  reduced,  the 
speed  is  reduced  momentarily,  and  a  later  cut-off  is  estab- 
lished by  the  springs  drawing  the  arms  toward  the  shaft  and 
changing  the  position  of  the  eccentric,  thus  bringing  the 
engine  back  to  speed. 


CHAPTER   XIII. 
SHAFT-GOVERNORS—ANALYSIS. 

THE  first  principle  of  shaft-governors — that  varying  the 
events  of  the  stroke  properly  will  cause  the  engine  to  run  at 
the  same  speed  under  all  loads — having  been  established,  the 
next  point  in  order  for  consideration  is  an  analysis  of  the 
action  of  the  springs  and  weights  whereby  the  movement  of 
the  eccentric  is  accomplished.  The  method  pursued  in  this 
will  be  one  drawn  from  a  valuable  monograph  entitled  "The 
Mechanics  of  the  Shaft-Governor,"  by  Prof.  Barr,  published 
in  the  Sibley  Journal  of  Engineering,  1 896. 

First,  reduce  the  springs  and  weights  to  the  simplest  form, 
shown  in  Fig.  88.  Here  there  is  but  one  weight,  represented 
by  the  ball  W^  which  slides  on  the  radial  rod  R.  The  length 
of  this  rod  is  such  that  when  the  ball  is  in  the  extreme  inner 
position  its  center  coincides  with  the  center  of  the  containing- 
wheel  or  spider  G.  The  movement  of  the  ball  is  resisted  by 
the  spring  5,  which  is  fastened  to  the  rim  of  the  spider.  The 
line  of  travel  of  the  spring  and  ball  is  a  diameter  of  the  circle. 
In  this  investigation  friction  is  neglected. 

First,  suppose  that  the  ball  is  moved  outward  by  any  force 
from  its  central  position  until  it  occupies  a  position  W.  In 
that  case  the  spring  must  be  extended  a  certain  amount,  equal 
to  the  distance  from  W^  to  W,  or  the  distance  the  ball  is 
moved;  and  it  will  then  exert  a  pull  on  the  ball,  tending  to 
restore  it  to  its  original  position.  The  amount  of  this  inward 
pull  is  found  as  follows.  "Spring  strength  "  is  the  term  used 
to  designate  the  force  in  pounds  required  to  extend  or  com- 

124 


SHAF7'-GO  VERNORS—ANAL  YSIS. 


I25 


press  a  spring  one  inch.  Thus,  a  2O-pound  spring  means  one 
which  requires  a  pull  of  20  pounds  to  extend  it  one  inch;  a 
6o-pound  spring  requires  a  pull  of  60  pounds  to  accomplish 
the  same  result.  The  total  amount  of  extension  or  compres- 


FIG.  88 

sion  is  directly  proportional  to  the  force  exerted  upon  the 
spring.  Thus,  a  weight  of  120  pounds  resting  on  the  20- 
pound  spring  would  compress  it 

120  -f-  20  =  6  inches, 

or,  if  directly  suspended  from  it,  would  lengthen  it  the  same 
amount.  The  same  weight  would  only  lengthen  or  shorten 
a  6o-pound  spring 

1 20  -T-  60  =  2  inches. 

The  weight  or  force  required  to  produce  any  given  change  in 
the  length  is  found  by  multiplying  the  spring  strength  by  the 


126  SLIDE-VAL  VES. 

amount  of  the  change  in  inches.  For  example,  to  lengthen 
a  3O-pound  spring  J  inch  would  require  a  force  of 

30  X  \  =  15  pounds. 

The  force  exerted  by  a  spring  is  equal  to  the  force  exerted  in 
compressing  or  extending  it  to  the  length  it  has.  The 
3O-pound  spring  just  mentioned  would  exert  a  force  of  15 
pounds  if  compressed  or  extended.  "To  every  action  there 
is  an  equal  and  opposite  reaction.'* 

Again,  suppose  the  wheel  to  revolve  with  the  ball  at  W. 
There  will  be  a  centrifugal  force  set  up  which  will  tend  to 
throw  the  ball  outward;  this  force  will  act  radially,  and  as  the 
ball  is  free  to  slide  along  the  rod  it  may  be  considered  that 
the  centrifugal  force  acts  along  that  line.  The  amount  of  this 
force  depends  upon  the  weight  of  the  ball,  the  rotative  speed 
of  the  wheel,  and  the  distance  from  the  ball  to  the  center  of 
the  wheel.  Expressed  in  a  formula,  it  is 

"  C=  .00002 84 WN*R,* ''"./.     .     .     (i) 

where  C  =  centrifugal  force  in  pounds; 
W=  weight  of  ball  in  pounds; 
N  =•  revolutions  per  minute; 

R  =  radius,  or  distance  from  the  center  of  the  wheel 
to  the  center  of  the  ball,  in  inches. 

This  may  be  stated  in  a  rule  as  follows: 

To  find  the  centrifugal  force  in  pounds  exerted  by  a  weight: 
Multiply  together  the  weight  in  pounds,  the  square  of  the  num- 
ber of  revolutions  per  minute,  and  the  radius,  or  distance  in 
inches  from  the  center  about  which  the  weight  revolves  to  the 
center  of  gravity  of  the  weight.  Multiply  this  product  by  the 
constant  .0000284,  and  the  result  is  the  centrifugal  force.  (In 
the  case  of  a  ball,  the  center  of  gravity  is  the  center  of  the 
ball.) 

For  example,    suppose    that    the   ball  at    W  weighs    20 


SHA FT- GO  VERNORS—ANA L  YSIS.  1 2 7 

pounds,  is  20  inches  from  the  center  of  the  wheel,  and  that 
the  latter  is  running  at  150  revolutions  per  minute.  The 
centrifugal  force  is  255.60  pounds,  found  by  the  above  rule  as 
follows:  The  square  of  the  number  of  revolutions  per  minute 
is  150  X  150  —  22,500.  Then  the  centrifugal  force  is 

C=  20  X  22500  X  20  X  .0000284  =  255.60  pounds. 

Formula  (i)  may  be  transposed  to  give  the  value  of  N 
when  the  other  factors  are  known,  giving 


(2) 


This  expressed  in  the  form  of  a  rule  is  as  follows: 

To  find  the  number  of  revolutions  per  minute  which  a  ball 
of  a  known  weight,  at  a  known  radius,  must  make  to  exert  a 
given  centrifugal  force  :  Multiply  the  weight  in  pounds  by  the 
radius  in  inches,  and  divide  the  centrifugal  force  by  this 
product.  Then  extract  the  square  root  of  the  quotient,  and  the 
figure  is  the  number  of  revolutions  per  minute. 

For  example,  if  a  weight  of  20  pounds  is  rotating  about 
a  center  20  inches  distant,  and  exerting  an  outward  pull  of 
255.60  pounds  on  the  link  connecting  it  with  that  center,  it 
is  making  150  revolutions  per  minute;  because,  applying 
formula  (2)  or  its  corresponding  rule, 

20  X  20  =  400 
255.60  -r-  400  =  .639 

^^639=  .7994 

187.7  x  .7994  =  150-04 

Now,  if  the  ball  when  running  at  any  given  speed  is  held 
stationary,  it  is  obvious  that  at  that  moment  the  centrifugal 
force  of  the  ball  and  the  spring  pull  must  be  equal  to  each 
other.  That  is,  taking  the  2O-pound  weight  which  was  just 
figured  to  have  255.60  pounds  of  centrifugal  force  urging  it 


128  SLIDE-  VA  L  VES. 

outward,  the  inward  spring  pull  must  be  255.60  pounds  in 
order  to  hold  the  ball  at  Wt.  It  is  obvious  that  this  amount 
of  spring  pull  could  be  secured  by  using  a  spring  of  any 
strength  and  stretching  it  the  required  amount;  or  the  stretch 
of  the  spring  could  be  assumed  and  the  strength  calculated. 
It  is  better  to  assume  the  spring  strength,  however,  because 
this  affects  the  closeness  of  regulation,  as  will  be  explained 
later. 

Suppose,  then,  that  the  spring  is  a  4<D-pound.  Then  to 
hold  the  ball  at  W  the  extension  necessary  is 

255.60  -^  40  =  6.39  inches. 

This  extension  will  not  permit  the  ball,  when  the  tension  is 
relieved,  to  move  into  the  center  of  the  containing-wheel; 
because  the  ball  is  20  inches  out  from  the  center,  so  that, 
with  the  tension  removed  and  the  spring  collapsed,  the  ball 
would  still  be 

20  —  6.39  =  13.61  inches 

away  from  the  center.  The  spring  pull  with  the  ball  that 
distance  out  from  the  center  would  be  zero,  and  therefore, 
to  maintain  a  balance  between  the  centrifugal  force  and  the 
spring  pull,  the  centrifugal  force  must  be  zero.  This  can 
only  happen  when  the  radius  at  which  the  weight  revolves,  or 
the  rotative  speed,  is  zero.  But  as  the  radius  is  13.61,  the 
rotative  speed  must  be  zero. 

Now  suppose  that  the  ball  moves  inward  to  Wz  ,  which  is 
4  inches  from  W,  or  16  inches  from  the  center.  In  that  case 
the  spring  extension  has  been  decreased  to  6.39  —  4  =  2.39 
inches,  and  the  spring  pull  to 

2.39  X  40  =  95.6  pounds. 

Then  the  speed  at  which  the  wheel  must  revolve  to  main- 
tain the  ball  at  W^ ,  the  revolutions  per  minute,  must  be  such 


SHA FT- GO  VERNORS—A NA L  YSIS.  1 29 

as  to  produce  a  centrifugal  force  of  95.6  pounds.     Formula 
(2)  or  Rule  II  gives  the  value  of  N  as  113  07. 


95-6 

2O  X  2O 


=  187.7  I/. 239 
=  113-67. 

That  is,  the  governor-ball  will  not  move  to  W^  until  the 
speed  has  decreased  to  113.67  revolutions  per  minute. 

Next,  find  the  effect  when  the  ball  is  on  the  other  side  of 
W,  say  at  W^  ,  which  is  4  inches  farther  out  than  Wy  or  24 
inches  from  the  shaft  center.  The  spring  extension  at  that 
place  is 

6-39  +  4=  10.39, 
and  the  spring  pull  is 

10.39  X  40  =  4I5-6. 

In  order  to  have  the  centrifugal  force  equal  to  the  spring 
pull,  the  revolutions  per  minute  must  increase  to  175.52, 
because,  from  formula  (2), 


-  :87.7     '    4'5'6 


20   X    24 


=  187.7  ^.86583 
=  187.7  x  .9351 
=  175.52. 

That  is,  with  the  engine  and  governor-wheel  running  at 
175.52  revolutions  per  minute  the  ball  would  be  held  in 
balance  at  W19  24  inches  from  the  center  of  the  shaft. 

Then  if  W9  and  W^  are  the  extreme  inner  and  outer  posi- 
tions of  the  governor-ball,  the  extreme  variation  in  speed  is 
from  113.67  revolutions  per  minute  to  175.52.  If  the  ball  in 


I 3°  SLIDE-  VA  L  VES. 

moving  from  W^  to  Wl  moves  the  governor  from  its  earliest 
to  its  latest  cut-off,  the  engine  will  vary 

175.52  -  113.67  =  61.85 

revolutions  per  minute  in  passing  from  light  to  full  load.     The 
normal  speed  of  the  engine  being  taken  at  150,  the  variation  is 

61.85 

-— -   =  41.23  percent. 

Prof.  Barr  has  devised  a  diagram  which  is  of  great  service 
in  showing  clearly  and  at  a  glance  the  relation  between  the 
spring  pull  and  centrifugal  force.  Fig.  89  shows  it  as  applied 
to  the  case  just  discussed. 

Draw  the  line  WQ  Wl ,  and  let  the  distances  on  it  from  W0, 
on  any  convenient  scale,  represent  distances  from  the  center 
of  the  shaft,  or  radii,  at  which  the  ball  revolves.  The  scale 
adopted  here  is  two  small  squares  to  the  inch.  Then  let  dis- 
tances perpendicular  to  W0W^  represent  the  forces  acting  on 
the  ball.  For  example,  it  was  shown  that  with  the  ball  at 
Wlt  the  spring  pull,  no  matter  what  the  speed,  is  415.6 
pounds.  Then  lay  off  W1S1  to  represent  this  amount.  The 
scale  chosen  in  the  figure  is  20  pounds  for  one  small  square; 
and  it  will  generally  be  found  necessary  to  employ  a  much 
smaller  scale  for  the  forces  than  the  distances  in  order  to  bring 
the  drawing  within  reasonable  limits.  Next,  lay  off  W^S^ 
equal  to  95.6,  which  was  found  to  be  the  spring  pull  with  the 
ball  at  W^ ,  or  16  inches  from  the  center.  Join  5,  and  52 ,  and 
produce  the  line  5,5,  until  it  meets  W0W,  at  50.  It  will  be 
found  that  S0  is  13.61  inches  from  W0.  This  is  as  it  should 
be;  for  it  was  shown  that  with  the  ball  at  that  distance  from 
the  center  the  spring  pull  is  zero.  The  line  505,,  represent- 
ing the  spring  pull,  is  straight,  because  the  pull  varies  directly 
as  the  extension  of  the  spring.  The  spring  pull  at  any  point 
is  then  found  by  drawing  a  perpendicular  from  the  point  on 
W^Wi ,  and  measuring  the  length  included  between  W^W^  and 


SHA FT-GO  VERNORS—A NA L  YSIS.  1 3 l 

the  line  of  spring  pull;  thus  with  the  ball  at  W,  20  inches 
from  the  center,  the  spring  pull  is  WS,  which  is  255.6  pounds. 

In  other  words,  horizontal  distances  from  50,  as  S^W^ 
S^W,  S^W^  ,  represent  the  spring  extension. 

Now  to  consider  the  centrifugal  force.  It  is  desirable  to 
have  the  engine  run  at  a  constant  speed,  and  the  centrifugal 
force  will  be  considered  on  that  basis.  There  must  be  some 
radius  at  which  the  spring  pull  and  the  centrifugal  force  at 
that  speed  will  balance.  For  example,  it  was  found  that  with 


the  engine  running  at  150  revolutions,  the  spring  pull  and  the 
centrifugal  force  balance  with  the  ball  at  W,  20  inches  out 
from  the  center.  Now,  with  a  constant  speed,  the  centrifugal 
force  varies  directly  as  the  radius,  as  shown  by  formula  (i). 
With  the  ball  at  the  center  of  the  wheel — -with  the  radius 
zero — the  centrifugal  force  is  zero.  Therefore  the  lines  of 
spring  pull  and  centrifugal  force  at  constant  speed  coincide  at 
20  inches  from  W^  and  WnCl  is  the  line  of  centrifugal  force, 
found  by  drawing  a  straight  line  from  W0  through  C.  The 
centrifugal  force  at  any  other  radius  is  found  by  drawing  the 


132  SLIDE-VALVES. 

perpendicular  from  the  end  of  the  radius  to  the  line  of  centrif- 
ugal force.  Thus  at  W^  the  centrifugal  force  is  WlCl ,  equal 
to  304.9  pounds,  and  with  the  ball  at  W^  the  centrifugal  force 
is  202.5  pounds,  measured  by  W^S^. 

Now  this  diagram  shows  very  clearly  that  if  the  speed  be 
maintained  constant,  the  centrifugal  force  and  the  spring  pull 
will  not  be  equal  at  different  radii.  They  are  equal  at  W 
only;  beyond  that  point  the  spring  pull  is  greater,  and  below 
it  the  centrifugal  force  is  the  larger.  This  shows  that  the 
governor  is  possessed  of  a  considerable  amount  of  stability; 
that  is,  it  is  but  little  liable  to  derangement  by  outside  forces, 
such  as  the  drag  of  the  valve.  Suppose,  for  example,  that 
the  engine  is  running  at  150  revolutions.  Then  the  ball 
balances  at  Wt  and  if  the  speed  be  maintained  constant,  a 
force  equal  to  S1Cl  must  be  applied  to  move  the  ball  out  to 
Wl  and  balance  it  there.  For,  if  the  ball  is  at  Wt,  the  cen- 
trifugal force  urging  it  outward  at  150  revolutions  is  Wf^ 
and  the  spring  pull  drawing  it  inwards  is  W^S^.  The  differ- 
ence between  these  two  is  C1S1 ,  equal  to 

415.6  —  304.9  =  1 10.7  pounds, 

which  is  the  force  necessary  to  send  the  ball  out  to  Wl  and 
hold  it  there.  Of  course,  if  a  lesser  force  were  applied,  it 
would  move  the  ball  somewhat,  but  it  would  not  force  it  out 
all  the  way. 

The  action  of  this  type  of  governor,  with  a  very  strong 
spring,  may  be  summed  up  as  follows:  It  does  not  give  very 
close  regulation,  the  example  chosen  showing  a  variation  of 
from  175  to  113  revolutions,  but  the  stability,  or  resistance  to 
external  forces,  is  very  great.  This  stability  increases  as  the 
spring  strength  increases,  which  shows  clearly  on  the  diagram, 
the  line  ss  being  the  spring  pull  for  the  stronger  spring.  This 
is  because  the  point  at  which  the  spring  pull  is  zero  is  at  a 
greater  distance  from  W0 ,  and  the  line  of  spring  pull,  passing 
through  5,  makes  a  greater  angle  with  the  line  of  centrifugal 


SHA  FT -GO  VERN  ORS—A  NA  L  YSIS. 


133 


force.  Hence  the  distance  between  the  two  lines  is  greater 
at  any  given  point;  and  as  this  distance  measures  the  force 
required  to  displace  the  ball,  the  stability  is  greater. 

Next,  consider  this  form  of  governor  having  in  a  spring 
of  such  strength  that  when  there  is  no  spring  extension  the 
ball  will  be  at  the  center  of  the  spider.  That  is,  with  the  ball 
out  anywhere  on  the  rod  the  spring  extension  is  equal  to  the 
radius  at  which  the  ball  revolves.  The  effect  of  this  spring 
can  be  best  understood  by  reference  to  Fig.  90,  which  is  the 


Barr  diagram  for  this  case.  Take  the  same  numerical  exam- 
ple as  before,  a  2O-pound  ball  running  at  150  revolutions  at  a 
radius  of  20  inches.  The  centrifugal  force  is  255.6  pounds  as 
determined  before,  and  the  spring  strength  must  be 

255.6  -v-  20  =   12.78  pounds 

to  bring  the  weight  to  the  center  of  the  spider  when  the 
spring  is  collapsed,  or  has  no  tension  on  it.  This  figure  is 
drawn  on  the  same  scale  as  the  preceding  one,  and  the  line  of 
centrifugal  force  at  constant  speed  is  the  same  as  before;  but 
the  line  of  spring  pull  is  decidedly  different.  At  W  the 
spring  pull  is  equal  to  the  centrifugal  force.  At  W0  it  is 
zero.  Therefore  the  straight  line  joining  these  two  points, 


134  SLWE-VAL  VES. 

which  represents  the  spring  pull,  coincides  with  the  line  of 
centrifugal  force. 

What  does  this  show  ?  It  shows  that  the  governor  regu- 
lates very  closely,  for  the  spring  pull  is  equal  to  the  centrif- 
ugal force  at  all  positions.  With  the  ball  out  at  Wl  the 
speed  would  be  150  revolutions,  and  with  the  ball  at  W^  the 
speed  would  be  the  same.  But  in  obtaining  this  closeness  of 
regulation  the  stability  has  been  sacrificed.  The  distance 
between  the  lines  of  spring  pull  and  centrifugal  force  measures 
the  resistance  to  deranging  forces;  and,  as  this  distance  is 
zero  at  all  points,  the  slightest  force  would  move  the  ball 
over  its  entire  length  of  travel.  The  slightest  increase  of 
speed  would  send  the  ball  out  to  its  extreme  outer  position, 
and  the  slightest  decrease  would  bring  it  in  to  the  center  of 
the  wheel.  A  governor  having  this  property  is  said  to  be 
isochronous. 

Fig.  91  shows  the  Barr  diagram  for  the  remaining  case  of 
the  elementary  governor,  that  in  which  the  spring  is  too  weak 
for  isochronous  action.  The  same  figures  are  used  as  before 
— weight  of  ball,  20  pounds,  revolutions  150,  spring  pull  and 
centrifugal  force  balancing  at  a  2O-inch  radius.  Assume  the 
spring  strength  as  8  pounds.  Then,  the  centrifugal  force  with 
W  at  20  inches  being  255.6  pounds,  the  spring  extension 
must  be 

255.6  -7-  8  =  31.95  inches. 
That  is,  the  spring  extension  is 

31.95  —  20=  11.95  inches 

greater  than  the  radius  at  which  the  ball  revolves.  This 
locates  the  point  SQ  over  at  the  left  of  W^ ,  as  shown  in  the 
figure.  Joining  S0  and  S,  which  is  located  the  same  as  in  the 
previous  cases,  gives  the  line  of  spring  pull.  Joining  W0  and 
5  gives  the  line  of  centrifugal  force  at  constant  speed.  This 
case  is  the  reverse  of  the  first  one,  shown  in  Fig.  89.  Beyond 


SHA FT-GO  VERNORS—A NA L  YSIS. 


135 


1 36  SLIDE-  VA  L  VES. 

W  the  centrifugal  force  is  greater  than  the  spring  pull,  and 
below  W  the  spring  pull  is  the  greater  of  the  two,  thus  show- 
ing that  this  form  of  governor  is  decidedly  unstable.  Sup- 
pose, for  example,  that  this  engine  is  running  at  its  normal 
speed  of  150  revolutions  with  the  ball  at  W.  Then  suppose 
a  part  of  the  load  to  be  suddenly  thrown  off.  As  a  natural 
consequence  the  engine  will  speed  up,  and  the  increased  cen- 
trifugal force  will  throw  the  ball  out ;  and  as  the  centrifugal 
force  increases  faster  than  the  spring  pull,  the  ball  will  go 
clear  out  to  its  extreme  position,  thereby  making  the  cut-off 
much  earlier  than  is  necessary  for  the  change  in  the  load. 
This  early  cut-off  of  course  results  in  a  reduction  of  speed. 
If  the  outer  limit  is  W^  where  the  spring  extension  is 

(31.95  +4)=  35-95  inches, 
and  the  spring  pull  is 

35-95  X  80  =  284.60  pounds, 

the  ball  will  balance  there  until  the  speed  is  reduced  below 
the  point  where  the  centrifugal  force  equals  2 84. 6  pounds,  or, 
from  formula  (2), 


N  = 


20  X    35-95 


=  I87.71/.3964 
=  187.7  X  .629 
=  1 1 8. 06  revolutions  per  minute. 

As  soon  as  the  speed  is  reduced  below  this  point  the  ball 
begins  to  move  inward;  and  as  the  centrifugal  force  decreases 
faster  than  the  spring  pull,  it  will  continue  to  move  in  until  it 
reaches  its  inner  limit.  This  will  result  in  a  later  cut-off  than 
is  necessary  to  bring  the  engine  back  to  speed,  and  the  result 
is  that  the  speed  will  grow  greater  than  is  necessary  to  hold 


SHA FT-GO  VERNORS—A NA L  YSIS.  1 3 7 

the  ball  at  W^.  If  this  inner  position  is,  as  before,  at  W9 , 
which  is  16  inches  from  the  center  of  the  wheel,  the  spring 
extension  at  that  point  will  be 

16  +  11.95  =  27.95  inches, 
and  the  spring  pull  will  be 

27.98  X  8  =  223.60. 

The  revolutions  required  to  produce  this  amount  of  centrif- 
ugal force  with  the  ball  out  16  inches  are 


=  I87V 


223.6 

20  X   16 


=  187/7^.69875 
=  187.7  X  .836 

=  156.92. 

That  is,  the  engine  must  speed  up  to  about  157  revolutions 
before  the  ball  will  move  out  again.  When  it  does  start,  it 
will  go  out  to  the  outer  end  again,  and  it  will  keep  on  "hunt- 
ing "  or  "racing  "  up  and  down,  the  speed  meanwhile  varying 
from  118  to  157  revolutions,  or 

(157-  118) 

^jp-  -;  =  26  per  cent. 

The  same  argument  holds  true  for  any  other  deranging 
force,  such  as  the  weight  of  the  valve  pulling  on  the  valve- 
rod,  so  that  this  form  of  governor  is  evidently  very  unstable. 

These  three  cases — stability,  isochronism,  and  unstability 
— cover  the  ground  completely.  The  first  two  qualities  are 
greatly  to  be  desired,  but  it  will  be  seen  at  once  from  the 
preceding  text  that  they  cannot  be  obtained  at  the  same 
time.  Stability  can  be  secured  easily  enough  by  using  a 
strong  spring,  but  this  renders  isochronism  out  of  the  ques- 


1 3  8  SLID E-  VA  L  VES. 

tion.  In  the  real  governor,  friction  in  the  moving  parts 
renders  the  governor  stable,  but  at  the  same  time  it  destroys 
the  desired  isochronism.  Good  results  have  been  secured  by 
reducing  the  external  forces  acting  on  the  governor  to  a 
minimum  and  then  making  the  governor  as  nearly  isochronous 
as  possible.  It  was  shown  that  a  perfectly  isochronous 
governor  is  a  possibility  when  the  spring  pull  and  centrifugal 
force  are  considered  to  be  the  only  forces  acting  on  the 
weight.  But  when  gravity  is  considered  it  will  be  found  that 
the  perfectly  frictionless  governor  will  race  between  its  outer 
and  inner  limits  under  a  steady  load.  This  is  due  to  the  fact 
that  when  the  ball  is  above  the  shaft,  gravity  tends  to  draw 
it  in,  and  when  the  ball  is  below  the  shaft,  gravity  tends  to 
draw  it  out,  or  away  from  the  center.  The  average  cut-off 
obtained  by  this  action  would  be  the  one  suited  to  the  load, 
but  would  be  alternately  too  early  and  too  late.  This  could, 
of  course,  be  obviated  by  making  the  friction  of  the  governor 
enough  to  prevent  the  racing,  but  this  is  a  poor  plan,  as  the 
desired  isochronism  is  thereby  rendered  impossible,  because, 
when  the  ball  moves  outward  to  compensate  for  a  reduction 
of  the  load  or  for  a  reduction  of  the  steam-pressure,  the 
centrifugal  force  must  overcome  the  friction  in  addition  to  the 
spring  pull.  The  best  method  of  overcoming  the  gravity 
distortion  is  to  balance  the  governor;  that  is,  to  employ  two 
weights,  or  their  equivalent,  one  on  each  side,  the  result 
being  that  the  gravity  effects  on  the  two  will  be  opposed  to 
each  other  and  will  therefore  be  negligible. 

The  same  argument  applies  to  the  disturbing  force  of  the 
valve;  it  could  be  compensated  for  by  friction  of  the  governor, 
but  it  is  not  desirable  to  so  arrange  it,  on  account  of  the 
destruction  of  isochronism.  With  an  ordinary  slide  valve 
this  pull  would  be  very  great,  because  the  steam  of  a  pressure 
equal  to  or  but  little  less  than  the  boiler-pressure  bears 
directly  upon  the  valve,  forcing  it  against  the  seat,  while  the 
pressure  underneath  the  valve  acting  upward  against  this  is 


SHA  FT-  G  O  VERNORS—A  NA  L  YSIS. 


'39 


only  that  of  the  exhaust  steam  which  is  expanded  to  a  con- 
siderably  lower  pressure.  This  unbalanced  downward  pressure 
increases  greatly  the  effort  required  to  move  the  valve. 
When  this  difference  of  pressure  is  obviated  or  greatly 
reduced,  the  valve  is  said  to  be  balanced. 

The    simplest    form    of    balanced    valve    is    the    piston- 


rfi 


FIG.  92. 

valve,  shown  in  section  in  Fig.  92.  The  valve  is  cylindrical, 
the  shape  being  similar  to  that  which  would  be  obtained  by 
rotating  a  plain  D  valve  about  its  valve-stem.  Consequently 
the  valve-seat  must  be  cylindrical  to  fit  the  valve-seat.  The 
valve  may  be  arranged  with  double  pistons,  such  as  in  Fig. 
92,  or  the  pistons  may  be  single,  as  in  Fig.  93.  By  referring 
to  either  figure,  it  will  be  seen  that  the  steajgi==f»^s*t«^  must 


140 


SLIDE-VALVES. 


be  equal  on  the  two  end  faces  of  the  valve ;  and  as  the  valve 
is  in  contact  with  the  seat  throughout  its  entire  length,  there 
is  no  possibility  of  any  other  pressure  than  friction  getting  at 
the  valve  sidewise.  The  objections  to  this  form  of  valve  are 
leakage  and  wear.  The  piston  cannot  fit  tight  in  its  bore, 
because  unequal  expansion  would  cause  the  piston  to  bind  in 
the  bore  when  the  latter  is  cold  and  the  pistons  hot.  This 
necessary  difference  in  size  would  result  in  leakage  if  not 


FIG.  93. 

guarded  against.  Spring  rings  are  used  to  effect  the  steam- 
tight  joint,  and  these  rings  cause  friction  and  wear.  Some- 
times no  rings  are  used,  but  in  that  case  great  care  must  be 
taken  to  keep  the  temperature  of  the  piston  and  the  bore  the 
same.  This  is  done  by  steam-jackets,  as  shown  in  Fig.  93. 

Another  system  of  balanced  valves  employs  pressure- 
plates.  Here  a  flat  plate  is  used  which  is  secured  in  the 
steam-chest,  and  which  receives  the  unbalanced  steam-press- 
ure, while  the  valve  slide  sunder  the  plate.  It  will  be  seen 
that  the  principle  is  practically  the  same  as  that  of  the  pistor  - 


SHAFT-GOVERNORS—ANAL  YSIS. 


141 


valve,  with  the  difference  that  with  the  pressure-plates  the 
valve  can  be  made  flat  and  steam-tight  at  the  top  and  bottom 
only,  where  it  touches  the  pressure-plate  and  seat. 

There  are  three  classes  of  pressure-plates — fixed,  adjust- 
able, and  flexible.  Fig.  94  shows  one  of  the  fixed  type. 
Here  the  plate  is  bolted  to  the  bottom  of  the  steam-chest,  and 
the  length  of  the  plate  is  such  that  the  valve  never  projects 


FIG.  94. 

beyond  it.  The  exposed  ends  of  the  valve,  being  of  equal 
area,  balance  each  other,  and  the  pressure  under  the  valve 
through  the  ports  is  balanced  by  recesses  of  equal  area  under 
the  hood  and  over  the  valve. 

In  some  cases  a  spring  ring  or  similar  arrangement  is  fitted 
to  the  top  of  the  valve  in  such  a  way  as  to  make  a  continuous 
contact  between  the  plate  and  the  valve,  thus  preventing  any 
steam  from  acting  on  the  top  of  the  valve.  Fig.  95  shows 
such  a  valve,  where  the  rings//  are  inserted  in  the  top  or 


142 


SLIDE-VALVES. 


back  of  the  valve  and  are  pressed  upward  against  the  plate  A 
by  means  of  springs.  The  space  between  the  plate  and  the 
back  of  the  valve  is  open  to  the  exhaust  through  the  opening 
//,  as  shown.  This  balances  nearly  all  the  top  of  the  valve, 
and  the  pressure  on  the  remaining  portion  is  sufficient  to  pre- 
vent leakage. 

An  example  of  the  second  type  is  shown  in  Fig.  96.  In 
this  design  the  pressure-plate  is  supported  on  an  inclined 
plane,  the  plate  being  made  to  slope  at  the  same  angle,  as 
shown  by  the  dotted  line  in  the  figure.  By  means  of  the 
adjustable  handle  E,  the  movable  plate  can  be  adjusted  to 


FIG.  95. 

secure  any  desired  amount  of  pressure  on  the  back  of  the 
valve.  The  valve  also  illustrates  another  type  of  double- 
ported  valve. 

The  third  system,  that  of  flexible  pressure-plates,  consists 
of  a  flexible  plate  of  steel  or  other  elastic  metal,  which  is  so 
arranged  as  to  allow  the  pressure  upon  it  to  force  it  down 
upon  the  valve,  but  only  with  force  enough  to  prevent  leak- 
age between  the  valve  and  the  plate.  An  example  of  this 
type  is  shown  in  Fig.  97. 

By  the  use  of  such  valves  as  these  the  friction  is  greatly 
reduced,  and  the  pull  on  the  governor  becomes  a  minimum. 

Another  disturbing  element  in  the  action  of  a  governor  is 
the  inertia  of  the  valve.  This  comes  into  play  at  each  end  of 
the  valve  travel,  when  it  is  necessary  to  reverse  the  direction 


. G  O  VERNORS—A  NA  L  YSIS. 


143 


144 


SLIDE-VALVES. 


of  motion  of  the  valve,  thus  giving  a  sudden  pull.  One 
method  of  obviating  or  compensating  this  deranging  force 
consists  of  employing  a  dash-pot.  This  consists  of  a  piston 
fitting  loosely  in  a  cylinder  containing  oil.  If  the  piston  is 
moved  slowly  and  steadily,  the  oil  offers  little  or  no  resistance 
to  the  motion,  as  it  will  readily  slip  past  the  circumference  of 
the  piston.  But  if  the  piston  be  moved  suddenly,  the  oil 
will  be  unable  to  flow  by  quickly  enough  to  permit  the  piston 
to  move  rapidly,  and  will  therefore  bank  up  and  impede  the 
progress  of  the  piston.  A  piston-rod  passes  through  a 


FIG.  97. 

stuffing-box  on  the  dash-pot,  and  is  attached  to  the  governor- 
weight,  the  dash-pot  itself  being  fastened  to  and  revolving 
with  the  governor-wheel  or  spider.  This  dash-pot  corrects 
the  distortion  due  to  the  inertia  of  the  valve,  and  in  no  way 
impairs  the  action  of  the  governor.  Another  result  is  obtained 
by  the  employment  of  this  contrivance.  When  the  weight 
starts  to  change  its  position  under  the  influence  of  a  change 
of  speed  it  starts  with  a  velocity  which,  owing  to  its  inertia, 
would  carry  it  beyond  the  proper  place  were  it  not  for  the 
retarding  influence  of  the  dash-pot.  The  dash-pot  then 
serves  as  a  preventive  of  too  sudden  a  motion  in  either  direc- 
tion. 


SHAFT-GO  VERNORS—ANAL  YSIS.  145 

The  last-mentioned  deranging  force  or  disturbing  element 
— the  inertia  of  the  governor-weight — may  be  turned  from  a 
hindrance  to  a  help,  making  it  perform  the  functions  of  a 
dash-pot.  This  is  explained  as  follows  by  Mr.  E.  J.  Arm- 
strong, in  a  paper  read  before  the  American  Society  of 
Mechanical  Engineers,  and  forming  a  part  of  Vol.  XI.  of  the 
Transactions :  When  any  governor  is  engaged  in  its  task  of 
controlling  the  engine,  the  weight  travels  at  a  variable  rate 
of  speed,  resulting  from  its  revolving  in  a  circle  of  variable 
size.  This  change  in  velocity  is  often  quite  considerable, 
depending  of  course  upon  the  amount  of  radial  movement 
and  the  rotative  speed.  To  take  an  example  from  the 
Straight-Line  engine:  a  fly-weight  is  i6£  inches  from  the 
center  of  the  shaft  when  in,  and  2oJ  inches  when  out,  mak- 
ing, at  220  revolutions  per  minute,  a  difference  of  194  feet  4 
inches  per  minute.  Whenever  the  fly-weight  takes  a  new 
position  it  must  change  its  speed — must  move  either  faster  or 
slower,  be  accelerated  or  retarded — and  must  absorb  or  give 
out  power  somewhere.  This  resistance  to  a  change  in  velocity 
acts  at  right  angles  to  a  radial  line  drawn  through  the  center 
of  gravity  of  the  weight,  and  if  the  weight  were  pivoted  so  as 
to  move  radially,  as  in  Fig.  98,  the  only  result  would  be  to 
increase  the  pressure  on  the  pivot.  If  the  fly-weight  were  so 
pivoted  as  to  move  at  an  angle  to  a  radial  line,  as  in  Fig.  99, 
so  that  in  its  outward  movement  it  goes  toward  the  way  the 
wheel  rotates,  then  the  outward  movement  of  the  fly-weight 
will  be  opposed,  to  some  extent,  by  this  resistance  to  accel- 
eration, depending  on  the  angle  which  the  line  of  movement 
forms  to  a  radial  line — or,  to  put  it  in  another  way — upon  the 
length  of  the  lever-arm  ABy  CA  being  the  line  of  resistance, 
and  B  the  fly-weight  pivot.  When  the  weight  moves  toward 
the  shaft,  the  action  is  the  same.  It  has  to  part  with  some 
of  its  momentum,  and  so  hangs  back,  as  in  its  outward  move- 
ment, thus  making,  similarly  to  a  dash-pot,  a  resistance  to 
movement  in  both  directions,  which  can  only  be  overcome 


146 


SLIDE-VALVES. 


FIG. 98 


FIG.  99 


SHAFT-GO  VERNORS—ANAL  YSIS. 


147 


quickly  by  a  great  force,  or  by  a  small  one  moving  slowly; 
this  resistance  increases  with  the  velocity  of  the  weight,  which 
is  all  that  is  accomplished  by  the  ordinary  dash-pot. 

To  return  to  the  analysis  of  the  governor  proper.  Fig. 
100  represents  a  slight  modification  of  the  elementary  type. 
The  change  consists  in  putting  a  stop,  T,  on  the  radial  rod, 
for  the  purpose  of  preventing  the  ball  from  traveling  in  to 
the  center  of  the  wheel.  With  the  wheel  at  rest  it  is  possible 


FIG.  100 

to  maintain  the  ball  in  the  position  shown  in  the  figure  by 
either  one  of  two  methods:  First,  by  the  employment  of  a 
spring  of  any  strength,  but  of  such  length  that  it  will  hold 
the  ball  in  that  position  without  any  extension — that  is,  the 
governor  will  be  of  the  stable  type,  such  as  explained  before, 
and  whose  action  is  illustrated  in  Fig.  89;  and,  second,  by 
shortening  such  a  spring,  and  then  extending  it  by  the  amount 


148  SLIDE-VALVES. 

that  it  is  shortened,  thus  getting  an  inward  pull  on  the  ball. 
This  pull  is  called  the  initial  tension,  because  it  is  the  tension 
existing  in  the  spring  while  the  wheel  is  at  rest,  or  before  any 
centrifugal  force  is  developed  to  extend  the  spring.  The 
amount  of  the  initial  tension  of  the  spring  may  be  expressed 
in  two  ways — the  amount,  in  inches,  that  the  spring  is 
shortened,  or  the  pull,  in  pounds,  on  the  ball.  Suppose  for 
example  that  a  lo-inch  spring  of  40  pounds  strength  will  hold 
the  ball  at  the  stop  without  any  extension.  If,  then,  this 
spring  is  shortened  to  7  inches  and  then  stretched  out  to  cover 
the  10,  the  pull  required  will  be 

(10  —  7)40  =  120  pounds, 

and  if  the  end  is  fastened  to  the  ball,  it  will  exert  that  pull  on 
it.  The  amount  of  the  initial  tension  in  pounds  is  therefore 
equal  to  the  amount  that  the  spring  is  extended  when  the 
wheel  is  at  rest,  and  the  ball  against  the  stop,  multiplied  by 
the  spring  strength.  If  the  initial  tension  is  expressed  by  the 
inches  that  the  spring  is  shortened,  the  spring  strength  must 
also  be  specified  in  order  to  render  it  exact.  In  the  practical 
governors  it  is  usual  to  supply  some  device,  such  as  a  screw- 
thread,  by  which  the  extension  of  the  spring  can  be  altered 
at  will,  and  any  desired  amount  of  initial  tension  secured. 

If  it  is  desired  to  produce  an  isochronous  governor,  it  is 
very  evident  that  the  initial  tension  must  be  such  that  if  the 
stop  on  the  radial  rod  were  removed,  the  ball  would  move 
inward  to  the  center  of  the  retaining  wheel. 

If  the  initial  tension  is  less  than  that  required  to  bring  the 
ball  in  to  the  center  of  the  wheel  with  the  stop  removed,  the 
result  will  be  the  same  as  if  a  strong  spring  were  employed ; 
that  is,  the  governor  will  be  stable. 

If  the  initial  tension  is  too  great  for  isochronism — that  is, 
if  the  ball  would,  on  removal  of  the  stop,  go  in  beyond  the 
center — the  result  obtained  would  be  the  same  as  if  a  weak 
spring  were  employed,  as  discussed  and  shown  in  Fig.  91. 


SHA FT-GO  VERNORS—A NA L  YSIS.  1 49 

This  initial  tension  produces  a  slight  variation  in  the 
operation  of  the  governor.  It  will  be  remembered  that  with 
the  elementary  isochronous  governor  the  weight  starts  out  as 
soon  as  the  engine  is  started;  but  with  initial  tension  the 
engine  must  continue  to  speed  up  until  it  is  going  so  fast  that 
the  centrifugal  force  of  the  ball  exceeds  the  initial  tension. 
For  example,  if  the  initial  tension  is  120  pounds,  and  if  the 
ball,  weighing  20  pounds,  is  held  by  the  top  at  the  point  6 
inches  from  the  center  of  the  wheel,  the  engine  will  speed  up 
until  the  centrifugal  force  is  greater  than  120  pounds;  that  is, 
from  from  the  rule  on  page  127,  until  the  revolutions  per 
minute  are  greater  than 

120    . 


=  187.7  Vi 

=  187.7. 

The  previous  argument  being  mastered,  it  is  next  in  order 
to  discuss  the  simple  type  of  shaft-governor  shown  in  Fig. 
101.  Here  the  weight  C  is  secured  to  the  lever  AC,  the  latter 
being  pivoted  at  A  on  an  arm  of  the  containing-wheel.  The 
spring  is  fastened  to  the  lever  at  B,  and  to  the  rim  of  the  con- 
taining-wheel at  D.  Now,  when  the  wheel  revolves,  a  certain 
centrifugal  force  is  induced  which  tends  to  throw  the  weight 
out,  that  is,  to  turn  the  lever  about  A.  This  is  resisted  by 
the  spring  as  before,  but  with  a  decided  modification.  The 
spring  pull  and  the  centrifugal  force  are  not  directly  opposed 
to  each  other,  but  act  on  the  lever  AC  at  different  distances 
from  A.  The  distance  AB  is  the  spring  leverage,  and  the 
distance  A  C  is  the  weight  leverage.  It  is  at  once  apparent 
that  in  order  to  secure  a  balance  between  the  opposing  forces 
at  any  given  speed,  it  must  be  true  that  spring  pull  X  spring 
leverage  =  centrifugal  force  X  weight  leverage.  Or,  putting 
this  in  symbols, 

5  X  L  =  C  X  L', 


1 5  O  SLIDE-  VA  L  VES. 

where  5  is  the  spring  pull; 

L  is  the  spring  leverage; 
C  is  the  centrifugal  force; 
L'  is  the  weight  leverage. 

Another  difference  between  this  and  the  elementary  type 
is  occasioned  by  the  fact  that  the  spring  and  weight  are  not 


FIG.  101 


directly  opposed  to  each  other;  that  is,  the  spring  extension 
is  not  equal  to  the  weight  displacement.  The  spring  exten- 
sion is  directly  proportional  to  its  leverage,  and  is  found  by 
multiplying  the  displacement  of  the  weight  by  the  spring 
leverage,  and  dividing  by  the  weight  leverage.  For  example, 
suppose  the  weight  leverage  to  be  8  inches,  while  the  spring 
leverage  is  but  4  inches,  and  that  the  weight  has  moved  out 


SHA FT-GO  VERNORS—A NA L  YSIS.  1 5 1 

6  inches  from  its  inner  position  under  the  influence  of  centrif- 
ugal force.     Then  the  spring  extension  is  equal  to 

6X4 

— - —  =  3  inches. 


That  is,  if  the  spring  has  half  the  leverage  of  the  weight, 
it  has  half  the  extension,  and  so  on.  It  must  be  borne  in 
mind  that  this  extension  is  from  the  inner  position  of  the 
weight.  If  a  stop  is  used  and  initial  tension  is  supplied,  the 
total  spring  extension  is  equal  to  that  above  found,  and  which 
may  be  called  centrifugal  extension,  increased  by  the  initial 
tension  in  inches. 

This  form  of  governor  possesses  certain  obvious  advan- 
tages. When  it  is  desired  to  secure  a  greater  spring  pull,  it 
is  not  necessary  to  substitute  a  stronger  spring.  The  same 
effect  may  be  secured  by  fastening  the  spring  at  a  greater  dis- 
tance from  A,  thereby  securing  a  greater  spring  leverage,  and 
at  the  same  time  increasing  any  existing  initial  tension,  or,  if 
none  exist,  adding  some. 

The  practical  governors  have  weights  of  various  shapes, 
but  the  principle  remains  the  same.  The  weight  is  supposed 
to  be  concentrated  at  its  center  of  gravity,  and  the  weight 
leverage  is  the  distance  from  this  center  of  gravity  to  the 
pivot,  irrespective  of  the  form  of  the  connecting  link.  Of 
course  the  weight  of  the  lever  must  be  taken  into  considera- 
tion in  finding  the  center  of  gravity.  The  radius  at  which  the 
centrifugal  force  is  found  is  the  distance  from  the  center  of 
gravity  to  the  center  of  the  containing-wheel. 

The  foregoing  text  leads  to  the  following  general  rules, 
which  are  applicable  to  all  governors  built  on  the  above 
principle;  and  this  type  of  governor  is  by  far  the  most 
common. 


1 5  2  SLIDE-  VA  L  VES. 

TO   INCREASE   THE    SPEED. 

It  may  be  desirable  to  increase  the  speed  at  which  the 
engine  is  to  run.  In  that  case,  increase  the  spring  tension, 
observing  the  following  limitations:  If  the  increase  of  spring 
tension  is  sufficient  to  produce  isochronism,  the  governor  will 
race  as  shown  before.  The  increase  must  be  kept  below  this 
limit,  and  as  the  governor  is  usually  set  near  this  limit,  the 
above  adjustment  is  applicable  only  to  small  changes  of  speed. 
For  larger  changes  the  governor-weight  may  be  decreased. 
This  will  produce  an  increase  of  speed,  because  the  centrifugal 
force  will  be  reduced  and  it  will  require  a  greater  number  of 
revolutions  per  minute  to  develop  sufficient  force  to  move  the 
weight  across  its  extreme  travel.  Or  the  increase  may  be 
secured  by  moving  the  weight  in  toward  the  pivot  about 
which  it  revolves.  This  lessens  the  weight  leverage,  and  con- 
sequently a  greater  number  of  revolutions  per  minute  are 
required  to  develop  the  same  centrifugal  force  as  before.  An 
increase  of  the  spring  leverage  will  produce  an  increase  of 
speed,  because,  the  spring  pull  being  thereby  increased,  a 
greater  centrifugal  force  is  required  to  stretch  it,  and  this  can 
only  be  secured  by  an  increase  in  the  number  of  revolutions 
per  minute. 

TO    DECREASE   THE    SPEED. 

The  adjustments  here  are  necessarily  the  opposite  of  those 
cited  above.  For  small  changes,  reduce  the  initial  tension. 
For  larger  changes,  the  weight  may  be  increased;  or  the 
weight  may  be  moved  out  from  the  pivot,  increasing  the 
weight  leverage;  or  the  spring  leverage  may  be  reduced. 

TO    REDUCE   THE   VARIATION    OF    SPEED. 

That  is,  to  produce  isochronism.     Increase  the   spring  ten- 
sion or  diminish  the  spring  leverage. 

All  of  these  changes  are  subject  to  one  limitation — that 


SHAFT-GO  VERNORS—ANAL  YSIS.  1 5 3 

of  performance.  It  is  impossible  to  set  a  governor  to  produce 
exact  results  by  theory  alone.  The  adjustment  must  be 
made  as  nearly  right  as  may  be,  and  then  the  engine  must  be 
run.  If  the  engine  runs  as  calculated,  all  well  and  good.  If 
not,  the  adjustment  must  be  carried  out  on  the  lines  indicated 
above  until  the  engine  runs  smoothly  and  regulates  closely. 


154 


SLIDE-VALVES. 


TABLE   I. 

VALUES   OF   O(y,  FIG.  2O,  FOR  VARIOUS    RATIOS   OF  CONNECTING-ROD   TO   CRANK. 


Ratio 

6>0'  =  Crank. 

Ratio 

OO'  =  Crank. 

Ratio 

OO'  =  Crank. 

of 
/-*_ 

of 
f-»  _ 

of 
/-.  _ 

Con- 
nect'g- 
rod  to 

Divided 

Multiplied 

Con- 
necting- 
rod  to 

Divided 

Multiplied 

Con- 
nect'g- 
rod  to 

Divided 

Multiplied 

Crank. 

by 

by 

Crank. 

by 

by 

Crank. 

by 

by 

I 

4 

.2500 

4 

16 

.0625 

7 

28 

.0356 

.1 

6 

.2000 
.1667 

3 

17 

.0588 
.0556 

7\ 

29 
30 

•0345 
•0333 

If 

7 

.1429 

4f 

19 

.0526 

7l 

31 

.0323 

2 

8 

.1250 

5 

20 

.0500 

8 

32 

•0313 

H 

9 

.1119 

If 

21 

.0476 

8i 

33 

.0303 

4 

10 

.lOOO 

sl 

22 

•0455 

8if 

34 

.0294 

2f 

ii 

.0909 

5t 

23 

•0435 

8f 

35 

.0286 

3 

12 

•0833 

6 

24 

.0417 

9t 

36 

.0278 

3? 

13 

.0769 

63- 

25 

.0400 

37 

.0270 

3* 

14 

.0714 

6£ 

26 

•0385 

9^ 

38 

.0263 

3f 

15 

.0667 

6f 

27 

.0370 

9t 

39 

.0256 

SHA  FT- G  O  VERNORS—A  NA  L  YSIS. 


TABLE   II. 

DECIMAL   EQUIVALENTS    OF    FRACTIONS    OF   AN   INCH. 


1/64 

.015625 

17/64 

.265625 

33/64 

•515625 

49/64 

.765625 

1/32 

•03135 

9/32 

.28125 

17/32 

.53125 

25/32 

.78125 

3/64 

.046875  ; 

19/64 

.296875 

35/64 

.546875 

51/64 

.796875 

1/16 

.0625 

5/i6 

.3125 

9/16 

•  5625 

13/16 

.8125 

5/64 

.078125 

21/64 

.328125 

37/64 

.578125 

53/64 

.828125 

3/32 

•09375 

17/32 

•34375 

19/32 

•59375 

27/32 

.84375 

7/64 

.109375 

23/64 

.359375 

39/64 

.609375 

55/64 

•859375 

1/8 

.125 

3/8 

.375 

5/8 

.625 

7/8 

.875 

9/64 

.  140625  ! 

25/64 

.390625 

41/64 

.640625 

57/64 

.890625 

5/32 

.15625 

13/32 

.40625 

21/32 

.65625 

29/32 

.90625 

11/64 

.171875 

27/64 

.421875 

43/64 

.671875 

59/64 

.921875 

3/i6 

•1875 

7/16 

•4375 

11/16 

.6875 

15/16 

•  9375 

13/64 

.203125  i 

29/64 

.453125 

45/64 

.703125 

61/64 

.953125 

7/32 

.21875 

15/32 

.46875 

23/32 

.71875 

31/32 

.96875 

15/64 

.234375 

31/64 

.484375 

47/64 

•734375 

63/64 

.984375 

i/4 

.25 

r/2 

•5 

3/4 

•75 

i 

.i 

I56 


SLIDE-VALVES. 


TABLE   III. 

EFFECT  OF  CHANGING  OUTSIDE  AND  INSIDE  LAPS,  TRAVEL  AND  ANGULAR 
ADVANCE.     (THURSTON.) 


CHANGE. 

ADMISSION. 

EXPANSION. 

EXHAUST. 

COMPRESSION. 

Increase 
Outside  Lap 

begins  later, 
ceases  sooner 

occurs  earlier, 
continues  longer 

unchanged 

unchanged 

Decrease 
Outside  Lap 

begins  earlier, 
ceases  later 

begins  later, 
period  shortened 

unchanged 

unchanged 

Increase 
Inside  Lap 

unchanged 

begins  as  before, 
continues  longer 

begins  later, 
ceases  earlier 

begins  sooner, 
contin.  longer 

Decrease 
Inside  Lap 

unchanged 

begins  as  before, 
period  shortened 

begins  earlier, 
ceases  later 

begins  later, 
period  short'd 

Increase 
Travel 

begins  sooner, 
ceases  later 

begins  later, 
ceases  sooner 

begins  later, 
ceases  later 

begins  later, 
ends  sooner 

Decrease 
Travel 

begins  later, 
ceases  earlier 

begins  earlier, 
ceases  later 

begins  earlier, 
ceases  earlier 

begins  earlier, 
ceases  later 

Increase 
Angular 
Advance 

begins  earlier, 
period 
unchanged 

begins  sooner, 
period 
unchanged 

begins  earlier, 
period 
unchanged 

begins  earlier, 
period 
unchanged 

Decrease 
Angular 
Advance 

begins  later, 
period 
unchanged 

begins  later, 
period 
unchanged 

begins  later, 
period 
unchanged 

begins  later, 
period 
unchanged 

SHAFT-GO  VERNORS—ANAL  YSIS. 
TABLE    IV. 

PORT   AREA,    PORT   WIDTH,    AND    STEAM-PIPE   DIAMETER   FOR   VARIOUS 
PISTON    SPEEDS   AND   STEAM   VELOCITIES. 


157 


Velocity  of  Steam.  Feet  per  Minute. 

4,000. 

6,000. 

8,000. 

10,000. 

12,000. 

4J 

3 

a 

G 
O 

V 

rt 

G 
0 

£ 

a 

c 

0 

§ 

3 

G 

o 

V 

rt 

1 

fc 

1 

rt 
g 

E 

I 

g 

E 

I 

g 

E 

O 

a 

.2 

8 

.2 

w 

s 

a 

tt 

£ 

to 

1 

rt 

8. 

< 

. 

Q 

^ 

w 

3 

**« 

u 

Q 

u 

3 

"^ 

C  . 

o 

G 
O 

S^1 

c 
o 

G 
O 

%  5 

c 

0 

I 

~.t? 

o 

§ 

iJ^ 

G 
0 

c 

<u  >, 

G 
O 

1 

fl 

.2 

E 

Is 

.a 

JJP 

tn 
E 

tn 

II 

.S3 

OH 

H 

S 

53 

38 

Q  IS 

Q""  tn 
i 

Q8 

1 

co 

G 

8  . 

a$ 

'II 

"O4J 

i. 

<  b 

m-pipe 
ameter 

jgS 

rt 
*  ' 

m-pipe 
ameter 

if 

cB 
QJ 

fl 

II 

d  . 

m-pipe 
ameter 

i! 

2 

? 

15 

CO 

!' 

|S 

So 
c/5 

|S 

1S 

co 

l] 

jjjo 

So 
c/5 

I3 

1S 

2  a 

V 

IOO 

.025 

.158 

.022 

.017 

.129 

.015 

.013 

.112 

.011 

.010 

.  IOO 

.009 

.008 

.091 

.007 

I25 

.031 
•037 

.177 

.194 

•  O27 

•  032 

.025 

^58 

.022 

.019 

.125 
•137 

.014 
.017 

.OI3 
•OI.S 

.123 

.013 

.013 

.112 

.009 

.Oil 

175 

.044 

.209 

.038 

.029 

.171 

•  025 

.022 

.I48 

.019 

.Ol8 

.132 

•  015 

.015 

.  121 

.013 

200 

.050 

.224 

.043 

•033 

•  183 

.029 

.025 

.158 

.022 

.020 

.141 

.017 

.017 

.129 

.015 

225 

.056 

•  237 

.049 

.038 

.194 

•°33 

.028 

.168 

.024 

.023 

.150 

.O2O 

.019 

•I37 

.016 

250 

.063 

.250 

•055 

.042 

.204 

•°37 

.031 

.177 

.027 

.025 

.158 

.022 

.021 

.144 

.018 

275 

.069 

.262 

.060 

.046 

.214 

.040 

•034 

.I85 

.030 

.028 

.166 

.024 

.023 

.020 

300 

•075 

•274 

.065 

.050 

.224 

.044 

.038 

•'93 

.033!  .030 

•  173 

.026 

.025 

•157 

.022 

325 

.081 

.285 

.070 

•054 

•233 

.047 

.041 

.201 

.036 

•033 

.180 

.028 

.027 

.164 

.024 

350 

.088 

.296 

.076 

.058 

.242 

.051 

.044 

.209 

.038 

•035 

.187 

.031 

.029 

.171 

.025 

375 

.094 

.306 

.O8l 

.063 

.250 

•055 

.047 

.2I7 

.041 

.038 

.194 

•033 

.031 

.177 

.027 

400 

.  IOO 

•3*3 

.086 

.067 

.258 

.050 

.224 

.044 

.040 

.200 

•035 

•033 

.183 

.029 

425 

.106 

.326 

.092 

.071 

.266 

.062 

•053 

.231 

.0461  .043 

.206 

•037 

•035 

.188 

•031 

45° 

•"3 

•335 

.098 

•075 

•274 

.065 

.056 

.238 

.049 

•045 

.212 

•°39 

.038 

•193 

.032 

475 

.119 

•344 

•  103 

.079 

.281 

.069 

•059 

.244 

.052 

.048 

.218 

.041 

.040 

.199 

•°33 

500 

•125 

353 

.108 

.083 

.288 

•073 

.250 

.055;  .050 

.224 

.044 

.042 

.204 

•036 

525 

•  362 

•"3 

.088 

•295 

.077 

!o66 

.256 

•058)  .053 

.220 

.046 

•  044 

.209 

.038 

550 

!ifj 

.119 

.092 

.302 

.080 

.069 

.262 

.060:  .055 

•235 

.048 

.046 

.214 

.040 

575 

.i44 

^380 

.124 

.096 

•309 

.084 

.072 

.268 

.063!  .058 

.240 

.050 

.048 

.219 

.042 

600 

.15° 

.388 

.130 

.100 

.316 

.087 

•075 

•274 

.o6s!  .060 

•245 

.052 

.050 

.224 

.044 

625 

.156 

•395 

•*35 

.104 

•323 

.091 

.078 

•279 

.068  .063 

.250 

•055 

.052 

.228 

•°45 

650 

.163 

•403 

.141 

.108 

•329 

.094 

.081 

•285 

.071 

•  065 

•255 

•057 

•  054 

•  232 

.047 

675 

.169 

.411 

.146 

•"3 

•335 

.098 

.084 

.290 

.074 

.068 

.260 

•059 

•  056 

•237 

.049 

700 

•  175 

.418 

.150 

.117 

•341 

.102 

.088 

.296 

.077 

.070 

-265 

.061 

.058 

.241 

.051 

725 

.181 

.426 

•155 

.121 

•347 

.106 

.091 

.301 

.079 

•073 

.269 

.063 

.060 

.246 

•°53 

750 

.188 

•433 

.161 

•  125 

•353 

.109 

.094 

.306 

.082  .075 

-274 

•065 

.063 

.250 

•°55 

775 

.194 

.440 

.166 

.129 

•359 

.097 

•3TI 

.085  .078 

.278 

.068 

.065 

•254 

•  056 

800 

.200 

•447 

.172 

•133 

•365 

!n6 

.100 

.316 

.087  .080 

.283 

.070 

.067 

•259 

.058 

825 

.206 

..177 

•137 

•371 

.120 

.103 

.321 

.090 

.083 

.287 

.072 

.0-59 

.262 

.060 

850 

•213 

.461 

.183 

.141 

•376 

.123 

.106 

•  326 

.093 

.085 

.292 

.074 

.071 

.266 

.062 

875 

.219 

.468 

.188 

•145 

.382 

.127 

.109 

.095 

.088 

.296 

.076 

.073 

.270 

.064 

900 

•225 

•474 

•  193 

.150 

-388 

.113 

^336 

.098 

.090 

.300 

.079 

•075 

.274 

•065 

925 

.231 

.481 

.198 

•154 

•393 

.134 

.116 

•340 

.101 

•093 

•3°4 

.O8l 

.077 

•277 

.067 

950 

.238 

-.487 

.204 

.158 

.398 

.138 

.119 

•344 

.104]  .095 

•308 

•083 

.079 

.281 

.069 

975 

.244 

•492 

.209 

.162 

•4°3 

.141 

•122   .349 

.106 

.098 

•  312 

•  085 

.081 

.285 

.071 

1000 

.250 

.500 

.214 

.166 

.408 

•MS 

•125   .353 

.109 

.100 

.316 

.087 

.083 

.289 

•073 

1025 

.256 

.506 

.220 

.170 

•413 

.149 

.128 

•357 

.112 

.103 

.320 

.089 

•  085 

.292 

1050 

•263 

.512 

.225 

•175 

.418 

•  153 

•13I 

.114 

.105 

•324 

.092 

.088 

•295 

.076 

I075 

.269 

.518 

•  231 

.179 

•423 

.156 

.134 

'365 

.117 

.108 

•  328 

.094 

.090 

.299 

.078 

IIOO 

.275 

•524 

•  236 

.183 

.428 

.160 

.138  .370 

.120 

.110 

•332 

.096 

.092 

•303 

.oSo 

1125 

.281 

•53° 

.241 

.187 

•433 

.163 

•J4i  -375 

.123 

•"3 

•335 

.098   '.094 

.306 

.082 

1150 

.288 

.536 

.246 

.191 

•438 

.167 

•144  -379 

.i26|  .115 

•339 

.100 

.096 

.310 

.084 

"75 

.294 

•542 

•  251 

•195 

•443 

.170 

•147  -384 

.128:  .Il8 

•343 

.103 

.098 

•3'3 

•  085 

1  200 

.300 

.548 

•  256 

.200 

•447 

•175 

.150  .388 

•I3t 

.120 

•346 

.105 

.IOO 

•  316 

.087 

i58 


SLIDE-VALVES. 


TABLE   V. 

AREAS    OF   CIRCLES. 
Advancing  by  Eighths. 


Diam. 

Area. 

Diam. 

Area. 

Diam. 

Area. 

Diam. 

Area. 

1/64 

.00019 

3 

7.0686 

u 

51.849 

207-39 

1/32 

.00077 

1/16 

7-3662 

8 

53-456 

•2 

210.60 

3/64 

•00173 

X£ 

7.6699 

% 

55-088 

x^ 

213.82 

1/16 

.00307 

3/i6 

7.9798 

% 

56.745 

% 

217.08 

.00690 

u 

8.2958 

% 

58.426 

94 

220.35 

X£ 

.01227 

5/i6 

8.6179 

% 

60.132 

7% 

223.65 

5/32 

.01917 

% 

8.9462 

7% 

61.862 

17 

226.98 

.02761 

7/16 

9  .  2806 

9 

63.617 

l£ 

230-33 

7/32 

•03758 

X<£ 

9.6211 

L£ 

65-397 

X4 

233-71 

M 

.04909 

9/16 

9.9678 

8 

67.201 

% 

237.10 

9/32 

.06213 

% 

10.321 

?8 

69.029 

X^ 

240.53 

.07670 

11/16 

10.680 

X£ 

70.882 

% 

243.98 

11/32 

.09281 

94 

11.045 

% 

72.760 

94 

247-45 

% 

.11045 

13/16 

11.416 

94 

74-662 

% 

250.95 

13/32 

.12962 

% 

"•793 

% 

76.589 

18 

7/16 

•15033 

15/16 

12.177 

10 

78.540 

258.02 

15/32 

•17257 

4 

12.566 

X£ 

80.516 

i^ 

261.59 

\/-£ 

•19635 

1/16 

12.962 

X^ 

82.516 

% 

265.18 

17/32 

.22166 

X£ 

13-364 

% 

84.541 

x^ 

268.80 

9/16 

.24850 

3/16 

I3-772 

1/2 

86.590 

% 

272.45 

.27688 

'H 

14.186 

% 

88.664 

94 

276.  12 

% 

.30680 

5/16 

14.607 

% 

90.763 

% 

279.81 

21/32 

.33824 

15-033 

/& 

92.886 

19 

283.53 

11/16 

.37122 

7/16 

15-466 

11 

95-033 

287.27 

23/32 

•40574 

/^ 

15.904 

X£ 

97.205 

/4 

291.04 

94 

.44179 

9/1  6 

'6.349 

/4 

99-402 

% 

294.83 

25/32 

•47937 

% 

16.800 

n 

101  .62 

X<j 

298.65 

13/16 

.51849 

11/16 

17.257 

n 

103.87 

% 

302.49 

27/32 

•559'4 

•M 

17.728 

rm 

106.14 

94 

% 

.60132 

13/16 

18.190 

%: 

108.43 

?2 

310.24 

29/32 

.64504 

% 

18.665 

% 

110.75 

20 

T 

15/16 

.69029 

15/16 

19.147 

12 

113.  10 

x^ 

318.10 

3V32 

•  73708 

5 

19-635 

x^ 

"5-47 

X4 

322.06 

1 

•7854 

1/16 

20.  129 

x^ 

117.86 

% 

326.05 

1/16 

.8866 

y 

20.629 

% 

120.28 

u 

330.06 

X^ 

•9940 

3A6 

2i.i35 

x^ 

122.72     . 

5^ 

334-io 

3/i6 

.1075 

N 

21.648 

% 

125.19 

94 

338.i6 

Y\6 

.2272 
•3530 

5/i6 

22.166 
22.691 

?8 

127.68 
130.19 

21 

346.36 

% 

.4849 

7/16 

23.221 

13 

x^ 

350.50 

7/16 

.6230 

n 

23-758 

x^ 

135.30 

H 

354-66 

Ni 

.7671 

9/16 

24.301 

X4 

137-89 

% 

358.84 

9/16 

•9175 

% 

24.850 

•Z 

140-50 

LA 

363-05 

% 

•°739 

11/16 

25.406 

x^ 

I43-I4 

% 

367  28 

11/16 

•2365 

N 

25.967 

% 

94 

371-54 

94 

•4°53 

13/16 

26.535 

% 

148.49 

% 

375  83 

13/16 
% 

.5802 
.7612 

'SA6 

27.109 
27.688 

14 

151.20 
153-94 

22 

380.13 
384.46 

15/16 

.9483 

6 

28.274 

x^ 

156.70 

H 

388.82 

2 

1/16 

3.1416 

H 

29-465 
30  .  680 

S 

159.48 
162.30 

H 

393  •  20 
397-6i 

X£ 

3*5466 

% 

31  -919 

x^ 

165.13 

% 

402.04 

3/'6 

/^ 

33-183 

% 

167.99 

94 

406  49 

X^ 

3  .9761 

!h6 

34-472 

94 

170.87 

/& 

410.97 

5/16 

4.2000 

% 

35.785 

% 

I73>78 

23 

415.48 

% 

4-4301 

% 

37.122 

15 

176.71 

x^ 

420.00 

7/16 

4.6664 

7 

38.485 

x^ 

179.67 

X4 

424-56 

L2 

4.9087 

/4 

39-87I 

x| 

182.65 

B 

429-13 

31 

5-I572 

% 

41.282 
42.718 

s 

185.66 
188.69 

H 

433-74 
438.36 

5-6727 

^ 

44.179 

% 

I9L75 

94 

443-or 

94 

5-9396 

% 

45.664 

94 

194.83 

% 

447.69 

13/16 

6.2126 

§4 

47-I73 

% 

^97  '93 

24 

452.39 

7/3 

6.4918 

fe2 

48.707 

16 

201.06 

IX 

4S7-" 

15/16 

6-7771       , 

8     ' 

50.265 

^ 

204.22 

..  ..J4 

461.86 

SHAFT-GOVERNORS—ANAL  YSlS. 


159 


AREAS    OF    CIRCLES. 


Diam. 

Area. 

Diam. 

Area. 

Diam. 

Area. 

Diam. 

Area. 

% 

466.64 

33 

855-3° 

% 

1360.8 

H 

1983.2 

% 

471.44 

j£ 

861.79 

% 

1369.0 

% 

% 

476.26 

/4 

868.31 

% 

1377-2 

t<£ 

2003.0 

¥4 

481.11 

% 

874-85 

42 

1385-4 

% 

2012.9 

H 

485.98 

12 

881.41 

/4 

1393-7 

% 

2022.8 

25 

490.87 

% 

888.00 

/4 

1402  .0 

% 

2032.8 

*A 

495-79 

% 

894.62 

% 

1410.3 

51 

2042.8 

% 

500.74 

% 

901.26 

Jl8 

1418.6 

i^ 

2052.8 

% 

505  -71 

34 

907.92 

% 

1427.0 

/4 

2062  .  9 

m 

510.71 

i^j 

914.61 

% 

M35  4 

78 

2073.0 

% 

5I5-72 

J4 

921.32 

% 

M43-8 

1^ 

2083.1 

% 

520.77 

•% 

928.06 

43 

1452.2 

98 

2093.2 

% 

525-84 

L<J 

934.82 

ly£ 

1460.7 

% 

2103.3 

26 

330-93 

% 

941.61 

}4 

1469  i 

% 

/^J 

536-05 

% 

948.42 

% 

1477.6 

52 

2123.7 

/4 

54I-19 

% 

955-25 

12 

1486.2 

2133-9 

% 

546.35 

35 

962  .  i  i 

% 

M94-7 

/4 

2144.2 

x^ 

551-55 

1  s 

969.00 

% 

% 

2154-5 

% 

556.76 

^4 

975-91 

% 

1511.9 

H 

2164.8 

% 

562.00 

% 

982.84 

44 

1520.5 

Kg 

2175.1 

% 

567.27 

C2 

989.80 

^& 

1529.2 

94 

2185.4 

27 

572.56 

% 

996-78 

/4 

1537-9 

% 

2195-8 

^ 

577.87 

% 

1003.8 

% 

1546.6 

53 

2206  .  2 

J4 

583-21 

% 

1010.8 

i  ., 

1555.3 

22IO.6 

% 

588.57 

36 

1017.9 

% 

1564-0 

/4 

2227.0 

zl3 

593  .  96 

i^ 

1025.0 

3^ 

1572.8 

&£ 

2237.5 

% 

599-37 

M 

1032.  i 

% 

1581.6 

IX 

2248.0 

M 

604.81 

fjj 

1039.2 

45 

1590.4 

% 

2258.5 

/^ 

610.27 

/^s 

1046.3 

^x 

J599-3 

% 

2269.1 

28 

6i5-75 

% 

JOSS-S 

14 

1608.2 

% 

2279.6 

^ 

621.26 

^4 

1060.7 

N 

1617.0 

54 

2290.2 

/4 

626.80 

?e 

1068.  o 

JX 

1626.0 

2300.8 

% 

632.36 

37 

1075.2 

% 

1634.9 

H 

/^ 

637-94 

/^ 

1082.5 

94 

1643.9 

'  % 

2322.1 

% 

643-55 

n 

1089.8 

% 

1652.9 

i^ 

2332.8 

M 

649-18 

% 

1097.1 

46 

1661.9 

•2 

2343-5 

% 

654.84 

Mi 

1104.5 

/^ 

1670.9 

^4 

2354.3 

29 

660.52 

% 

mi  .8 

/4 

1680.0 

% 

23^5.0 

^ 

666.23 

% 

1119.2 

sx 

1689.1 

55 

2375-8 

^4 

671.  g6 

% 

1126.7 

L^ 

1698.2 

2386.6 

9i 

677.71 

38 

1134.1 

% 

1707.4 

H 

2397-5 

lit 

683.49 

^ 

1141.6 

% 

1716.5 

^ 

2408.3 

It 

689.30 
695-13 

% 

1149.1 
1156.6 

47  ^ 

1725-7 
1734-9 

K 

2419.2 
2430.1 

n.     % 

700.98 

i^j 

1164.2 

/^ 

1744.2 

34 

2441.1 

30 

706.86 

% 

1171.7 

/4 

J753-5 

% 

2452.0 

M 

712.76 
718.69 

8 

"79-3 
1186.9 

ft 

1762.7 
1772.1 

56 

2463.0 
2474.0 

94 

724-64 

39 

1194.6 

5,g 

1781.4 

/4 

2485.0 

/-ia 

730.62 

/^ 

1202.3 

§4 

1790.8 

% 

2496.1 

/^& 

736.62 

i^ 

121O.O 

TA 

1800.1 

iz 

2507.2 

% 

742.64 

?1 

1217.7 

48  j| 

1809.6 

^| 

2518.3 

n*      /" 

748.69 

U 

1225.4 

1819.0 

3£ 

2529.4 

31 

754-77 

% 

1233.2 

14 

1828  5 

% 

2540.6 

/^ 

760.87 

^4 

I24I.O 

% 

1837.9 

57 

255L8 

/4 

766.99 

?i 

1248.8 

t2 

1847-5 

2563.0 

% 

773-M 

40 

1256.6 

% 

1857.0 

J4 

2574-2 

J^ 

779  3i 

J4 

1264.5 

M 

1866.5 

§1 

2585.4 

/o 

785-51 

/4 

1272.4 

^« 

1876.1 

IX 

2596.7 

$4 

791.73 

12 

1280.3 

49. 

1885.7 

% 

2608  .  o 

32  ^ 

804.25 

H 

1288.2 
1296.2 

H 

1895.4 
1905.0 

% 

2619.4 
2630.7 

J^J 

810.54 

^4 

1304.2 

N 

1914.7 

58 

2642.1 

^4 

816.86 

7^6 

I3I2.2 

vz 

1924.4 

1^ 

26"  ->     r 

g 

823.21 
829.58 

41 

1320.3 
1328.3 

1 

1934.2 
1943-9 

N 

2664.9 
2676.4 

% 

835-97 

^4 

1336.4 

% 

1953-7 

l^j 

2687.8       . 

% 

842.39 

9i 

1344.5 

50 

^ 

2699.3 

* 

848.83 

JJ 

1352.7 

J5 

1973-3 

N 

2710.9 

i6o 


SLIDE-VALVES, 


AREAS    OF    CIRCLES. 


Diam. 

Area. 

Diam. 

Area. 

Diam. 

Area. 

Diam. 

Area. 

y 

2722.4 

34 

3578.5 

H 

4556.4 

M 

5641.2 

59 

2734-0 

% 

359x-7 

34 

4565-4 

% 

5657-8 

2745-6 

% 

3605.0 

% 

4581-3 

85 

5674-5 

n 

2757.2 
2768.8 

68 

3618.3 
3631  '7 

H 

4596.3 
4611.4 

^ 

5691-2 
5707-9 

34 

2780.5 

3645-0 

M 

4626.4 

% 

5724-7 

% 

2792.2 

34 

3658.4 

% 

4641.5 

34 

574T-5 

H 

2803.9 

% 

3671.8 

77 

4656.6 

% 

5758.3 

% 

2815.7 

34 

3685-3 

34 

4671.8 

M 

5775-1 

60 

2827.4 

% 

3698.7 

34 

4686.9 

/& 

579I-9 

2839.2 

% 

3712.2 

% 

4702.1 

86 

5808.8 

34 

2851.0 

% 

3725-7 

34 

47T7-3 

34 

5825.7 

a2 

2862.9 

69 

3739-3 

% 

4732-5 

34 

5842.6 

iz 

2874.8 

34 

3752-8 

M 

4747-8 

% 

5859-6 

% 

2886.6 

H 

3766.4 

% 

4763  -1 

34 

5876.5 

M 

2898.6 

% 

3780.0 

78 

4778.4 

% 

5893.5 

T^j 

2910.5 

34 

3793-7 

34 

4793-7 

3^i 

5910.6 

61 

2922.5 

% 

34 

4809.0 

% 

5927.6 

2934-5 

M 

3821.0 

N 

4824.4 

87 

5944-7 

34 

2946-5 

7^3 

3334.7 

34 

4839-8 

34 

5961.8 

£2 

2958-5 

70 

3848-5 

% 

4855-2 

34 

5978.9 

34 

2970.6 

34 

3862.2 

^4 

4870.7 

% 

5996.0 

% 

2982.7 

H 

3876.0 

/•s 

4886.2 

34 

6013.2 

% 

29  H-  8 

^1 

3889.8 

79 

4901-7 

% 

6030.4 

% 

3006.9 

34 

3903.6 

/4 

4917.2 

% 

6047.6 

62 

3019-1 

% 

39J7.5 

34 

4932-7 

% 

6064.9 

i  ^ 

3Q31  -3 

§4 

3931-4 

% 

4948.3 

88 

6082.1 

34 

3043.5 

/-i 

3945-3 

34 

4963-9 

34 

6099.4 

•2 

3055.7 

71 

3959-2 

% 

4979-5 

34 

6116.7 

i<£ 

3068.0 

34 

3973  -1 

M 

4995-2 

% 

6134.1 

^8 

3080.3 

34 

3987.1 

^8 

5010.9 

34 

6i=;i  .4 

M 

3092  6 

% 

4001.1 

80 

5026.5 

% 

6168.  8 

vb 

3104.  9  f 

34 

4015    2 

34 

5042-3 

% 

6186.2 

63 

3117.2* 

% 

4029.2 

H 

5058.0 

% 

6203.7 

i^ 

3129.6 

§4 

4043.3 

% 

5073-8 

89 

6221.1 

H 

3142.0 

/•i 

4057.4 

34 

5089.6 

34 

6238.6 

8 

3I54-5 

72 

407L5 

% 

5105.4 

3^ 

6256.  i 

34 

3166.9 

34 

4085.7 

M 

5121.2 

% 

6273.7 

^6 

3I79-4 

34 

4099.8 

?8 

5I37-1 

34 

6291.2 

3X 

9& 

4114.0 

81 

S^.o 

•Z 

6308  .  8 

% 

3204-4 

34 

4128.2 

34 

5168.9 

M 

6326.4 

64 

3217.0 

fa 

4142.5 

34 

5184.9 

% 

6344.1 

3229.6 

% 

4156.8 

Q 

5200.8 

90 

6361.7 

34 

3242-2 

% 

4171.1 

34 

5216.8 

34 

6379-4 

a2 

73 

4185.4 

% 

5232.8 

34 

6397  -1 

34 

3267-5 

34 

4I99.7 

M 

5248.9 

% 

6414.9 

% 

3280.1 

3^ 

4214.1 

% 

5264.9 

34 

6432.6 

% 

3292.8 

% 

4228.5 

82 

5281  .0 

% 

6450.4 

7^2 

3305.6 

34 

4242.9 

34 

5297.1 

$4 

6468.2 

65 

3318.3 

M 

34 

53J3-3 

% 

6486.0- 

AX 

4271.8 

3z 

91 

6503.9 

34 

3343-9 

% 

4286.3 

34 

5345-6 

34 

6521.8 

a2 

3356-7 

74 

4300.8 

% 

536t.8 

34 

6539-7 

H 

3369.6 

34 

43I5-4 

§4 

5378.1 

?i 

6557-6 

M 

3382-4 

H 

4329-9 

% 

5394-3 

34 

6575-5 

a£ 

3395-3 

% 

4344-5 

83 

5410.6 

% 

6593-5 

% 

3408.2 

34 

4359-  2 

34 

5426.9 

% 

6611.5 

66 

3421.2 
3434-2 
3447  -  2 

s 

4373-8 
4388.5 
4403.1 

1 

5443-3 
5459-6 
5476.0 

92  ^ 

6629  .  6 
6647.6 
6665.7 

% 

3460.2 

75 

4417.9 

% 

5492.4 

34 

6683.8 

34 

3473  2 

34 

4432-6 

% 

5508.8 

% 

6701  .9 

M 

3486.3 

34 

4447-4 

/& 

5525-3 

V^ 

6720.1 

94 

3499-4 

n 

4462.2 

84 

5541-8 

% 

6738.2 

% 

3512.5 

34 

4477-0 

34 

5558.3 

8 

6756.4 

67 

3525-7 

% 

4491.8 

M 

5574-8 

?» 

6774-7 

l^C 

3538.8 

§4 

4506.7 

•Z 

559^4 

93 

6792.9 

\A 

3552.0 

7Z 

i^ 

5607.9 

1^2 

6811.2 

% 

3565-2 

76 

4536.5 

" 

5624.5 

34 

6829.5 

SHA  FT-  G  O  VERNORS—A  NA  L  YSIS. 


161 


AREAS    OF   CIRCLES. 


Diatn. 


Area. 


94 


95 


6847-8 
6866.1 
6884.5 
6902.9 
6921.3 
6939.8 

6958 . 2 
6976.7 

6995-3 
7013.8 
7032.4 
7051.0 
7069.6 
7088.2 


Diam. 


96 


Area. 


7106.9 
7125.6 

7M4-3 
7163.0 
7181.8 
7200.6 
7219.4 
7238.2 
7257 -1 
7276.0 

7313.8 
7332  8 
7351-8 


Diam. 


98 


Area. 


737° -8 
7389.8 
7408.9 
7428.0 
7447 -1 
7466.2 

7504-5 
7523-7 
7543-° 
7562.2 

758i.5 
7600.8 
7620.1 


Diam. 


99 


100 


Area. 


7639.5 
7658.9 
7678.3 
7697.7 
7717.1 
7736.6 
7756.1 

7775-6 
7795-2 
7814.8 

7854.0 


INDEX. 


Admission 32 

Advance,  Angle  of . 12 

Advancing  Eccentric,  Effect  of 41 

Allen  Valve 82 

Angle  of  Advance 85. 

Inside  Lap 83 

Laying  out  False  Seat ;  88 

Maximum  Port-opening 85 

Outside  Lap 85 

Port  Area 84 

Port  Width.. 84 

Valve-travel 85 

Analysis  of  Shaft-governors 124 

Angle  of  Advance 13,  14 

Determination  of 56,  58,  59 

Effect  of  Changing 41 

A  ngle  between  Crank  and  Eccentric 13 

Angle  of  Follow 14 

Angular  Advance.      See  Angle  of  Advance. 

Angularity  of  Connecting-rod 19 

Eccentric-rod 22 

Balanced  Valves 130; 

Barr  Diagram 62 

Bridge 12 

Width  of 5I 

Buckeye  Governor b 123 

Center,  Dead 6 

Placing  Engine  on 100 

Centrifugal  Force 126 

Changing  Angular  Advance,  Effect  of 41 

Dimensions  of  Valve,  Effect  of 41 

Inside  Lap,  Effect  of 42 

163 


164  INDEX. 

PAGE 

Changing  Outside  Lap,  Effect  of 41 

Valve-travel,  Effect  of 41 

Compression 8,  34 

and  Exhaust,  Equalization  of 67 

Connecting-rod 17 

Angularity  of 19 

Infinite 6 

Construction  of  Eccentric 1 8 

Crank.. 17 

Crank  Angle 10 

Crank  and  Connecting-rod 17 

Crank-end 1 1 

Crank-pin 6,  17 

Crank  and  Piston  Positions,  Relative 23 

Crank-shaft 6,  17 

Crank,  Throw  of 8 

Cross-head 17 

Cushioning 8 

Cut-off 8,  32 

and  Lead,  Equalization  of 65 

Dash-pot 144 

Dead-center 6 

Placing  Engine  on 100 

Decreasing  Speed  of  Engine. , 152 

Variation 152 

Deprez  Method  for  Relative  Crank  and  Piston  Positions 23 

Design  of  an  Allen  Valve. .... 82 

a  D-valve 68 

a  Double-ported  Valve 90 

a  Trick-valve 82 

General  Problems  of 54 

Diagram,  Barr 130 

Valve 27 

Diameter  of  Exhaust-pipes 48 

Steam-pipes > 48 

Dimensions  of  Exhaust-port 48 

Steam-port 47 

Valve,  Effect  of  Changing 41 

Displacement  of  Valve 10,  14,  27 

Drag  of  Valve. 138 

D-valve  Design 68 

Angular  Advance 69 

Bridge 74 

Exhaust-port..   74 


INDEX.  165 

PAGE 

D-valve  Design,  Inside  Lap 72 

Laying  out  Valve 76 

Maximum  Port-opening 68 

Outside  Lap 69 

Port  Area 68 

Port  Width   . .    , 68 

Valve-travel 74 

Double-ported  Valve  Design 9o 

Bridge 95 

Exhaust-port 95 

Laying  out  the  Valve 97 

Maximum  Port- opening  ; 94 

Port  Area . . . 94 

Port  Width 94 


Eccentric,  Construction  of 18 

Following  Crank 14 

Leading  Crank 14 

Eccentric-pin 7 

Eccentric-rod 7 

Angularity  of 22 

Length  of 80 

Eccentric,  Shifting 113,  us 

Swinging 113,114 

Throw  of 8 

Eccentricity 8 

Determination  of 58,  60 

Effect  of  Changing  Angle  of  Advance 41 

Dimensions  of  Valve 41 

Inside  Lap 42 

Outside  Lap 41 

Valve-travel   41 

Effect  of  Inside  Lap 12 

Outside  Lap 12 

Equal  Cut-off,  Valve-setting  for . .    104 

Lead,  Valve-setting  for  102 

Equalizing  Cut-off  and  Lead 65 

Exhaust  and  Compression 67 

Exhaust-lap 12 

Exhaust-port 3 

Dimensions  of 48 

Maximum  Opening 49 

Exhaust-pipes,  Diameter  of 48 

Expansion  of  Steam,  Economy  of 8 


1 66  INDEX. 

PAGE 

Governors,  Shaft 108 

Analysis 124 

Buckeye 123 

Diagram 130 

Isochronous 133 

Stable... 131 

Straight-Line 122 

Unstable 134 

Weight,  Inertia  of 144 

Westinghouse 122 

Guides 17 

Head-end 1 1 

Increasing  Speed  of  Engine 152 

Inertia  of  Governor-weight 144 

Valve 143 

Infinite  Connecting-rod 6 

Initial  Tension 148 

Inside  Lap 1 1 

with  Infinite  Rod 1 1 

Effect  of  Changing 42 

Isochronous  Governor. 133,  152 

Lap,  Determination  of 54>  5$>  59 

Effect  of 12 

Inside n 

Outside 9 

Lead 4° 

Lead-angle 40 

Lead  and  Cut-off,  Equalization  of 65 

Determination  of. 54>  5$ 

Length  of  Eccentric-rod 80 

Port 47 

Valve-chest 7& 

Valve-stem 79 

Maximum  Port-opening 49 

for  Exhaust 49 

Outside  Lap. 9 

Effect  of 12 

with  Infinite  Rod 10 

Pipes,  Diameter  of 48 


INDEX.  167 

PAGE. 

Piston  and  Crank  Positions 23 

Piston-rod 3 

Piston-speed 68,  84,  93 

Piston-valve 139 

Pitman '•'•"• . . . 17 

Ports 2 

Area  of . . . . . 44,  68 

Length  of 47.  48 

Width 47,68 

Port-opening 27 

Exhaust 33 

Maximum 49 

Problems  of  Design 54 

Reducing  Speed  of  Engine 152 

Variation 152 

Release 34 

Reversing  Engine 14,  108 

Rockers 7 

and  Bell  Cranks 62 

Running  "  Over  " 15 

"Under" 16 

Setting  Valves 98 

with  Chest-cover  on 106 

for  Equal  Cut-off 104 

Leads 102 

Shaft-governors.     See  Governors. 

Shifting  Eccentrics 113,  118 

Slide-valve,  Design  of  a  D 68 

No  Laps 2 

Slotted  Cross-head  , 6 

Speed,  Decreasing 152 

Increasing 152 

Variation,  Decreasing 152 

Spring  Extension 125 

Strength 1 24 

Stable  Governor 131 

Steam-chest 2,  78 

Steam-pipes ; 48 

Steam-ports 2 

Dimensions 47 

Maximum  Opening 49 

Steam,  Velocity  of 44 

Straight- Line  Governor 122 


1 68  INDEX. 

t 

PAGE 

Stuffing-box 3 

Swinging  Eccentrics 113,  114 

Trick- valve.     See  Allen  Valve. 

Throw  of  Crank  and  Eccentric 8 

Unstable  Governor 134 

Valve,  Allen.      See  Allen  Valve. 

Balanced 139 

D-.     See  D-valve. 

Double-ported.      See  Double-ported  Valve. 

Trick-.     See  Allen  Valve. 

Valve-chest,  Length  of 78 

Valve-circle 27 

'Valve  Design,  General  Problems 54 

Examples 68,  82,  90 

Valve-diagrams 27 

Valve-displacement 10,  14,  27 

Valve,  Drag  of 138 

Inertia  of 143 

Piston- 139 

and  Piston,  Relation  between 5 

Valve-rod 3 

Valve-seat 3 

Valve-setting 99 

with  Chest-cover  on 106 

for  Equal  Cut-offs. .  ^ 104 

Leads 102 

Valve-stem,  Length  of 79 

Valve-travel , 8 

Determination  of 56,  59 

Effect  of  Changing 41 

Variation  of  Speed,  Reducing 152 

Velocity  of  Steam. 44 

Westinghouse  Governor 51 

Wrist-pin 17 

Zeuner  Diagram  27 


I 


'NIVERSITY  OF  CALIFORNIA  LIBRARY 
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iration  of  loan  period. 


DEC  20  1916 
17 1917O) 
31 191 A  V 


001271988 


50?n-7,'16 


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